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Page No 18.11:

Question 1:

Using binomial theorem, write down the expansions of the following:
(i) 2x+3y5

(ii) 2x-3y4

(iii) x-1x6

(iv) 1-3x7

(v) ax-bx6

(vi) xa-ax6

(vii) x3-a36

(viii) 1+2x-3x25

(ix) x+1-1x

(x) 1-2x+3x23

Answer:

(i) (2x + 3y)5

=C05(2x)5(3y)0+C15(2x)4(3y)1+C25(2x)3(3y)2+C35(2x)2(3y)3+C45(2x)1(3y)4+C55(2x)0(3y)5
=32x5+5×16x4×3y+10×8x3×9y2+10×4x2×27y3+5×2x×81y4+243y5=32x5+240x4y+720x3y2+1080x2y3+810xy4+243y5

(ii) (2x − 3y)4

=C04(2x)4(3y)0-C14(2x)3(3y)1+C24(2x)2(3y)2-C34(2x)1(3y)3+C44(2x)0(3y)4=16x4-4×8x3×3y+6×4x2×9y2-4×2x×27y3+81y4=16x4-96x3y+216x2y2-216xy3+81y4

(iii)
x-1x6=C06 x61x0-C16 x51x1+C26 x41x2-C36 x31x3+C46 x21x4-6C5 x11x5+C66 x01x6=x6-6 x5×1x+15 x4×1x2-20x3×1x3+15x2×1x4-6 x×1x5+1x6=x6-6x4+15x2-20+15x2-6x4+1x6

(iv) (1 − 3x)7
=C07(3x)0-C17(3x)1+C27(3x)2-C37(3x)3+C47(3x)4-C57(3x)5+C67(3x)6-C77(3x)7=1-7×3x+21×9x2-35×27x3+35×81x4-21×243x5+7×729x6-2187x7=1-21x+189x2-945x3+2835x4-5103x5+5103x6-2187x7

(v) 

(ax-bx)6=C06(ax)6(bx)0-C16(ax)5(bx)1+C26(ax)4(bx)2-C36(ax)3(bx)3+C46(ax)2(bx)4-C56(ax)1(bx)5+C66(ax)0(bx)6
=a6x6-6a5x5×bx+15a4x4×b2x2-20a3b3×b3x3+15a2x2×b4x4-6ax×b5x5+b6x6=a6x6-6a5x4b+15a4x2b2-20a3b3+15a2b4x2-6ab5x4+b6x6

(vi)
xa-ax6=C06xa6ax0-C16xa5ax1+C26xa4ax2-C36xa3ax3+C46xa2ax4-C56xa1ax5+C66xa0ax6=x3a3-6x2a2+15xa-20+15ax-6a2x2+a3x3

(vii)
x3-a36=C06(x3)6(a3)0-C16(x3)5(a3)1+C26(x3)4(a3)2-C36(x3)3(a3)3+C46(x3)2(a3)4-C56(x3)1(a3)5+C66(x3)0(a3)6=x2-6x5/3a1/3+15x4/3a2/3-20xa+15x2/3a4/3-6x1/3a5/3+a2


(viii)
(1+2x-3x2)5Consider 1-2x and 3x2 as two separate entities and apply the binomial theorem.Now,C05(1+2x)5(3x)0-C15(1+2x)4(3x2)1+C25(1+2x)3(3x2)2-C35(1+2x)2(3x2)3+C45(1+2x)1(3x2)4-C55(1+2x)0(3x2)5=(1+2x)5-5(1+2x)4×3x2+10×(1+2x)3×9x4-10×(1+2x)2×27x6+5(1+2x)×81x8-243x10=C05×(2x)0+C15×(2x)1+C25×(2x)2+C35×(2x)3+C45×(2x)4+C55×(2x)5-     15x2[C04(2x)0+C14(2x)1+C24(2x)2+C34(2x)3+C44(2x)4]+      90x4[1+8x3+6x+12x2]-270x6(1+4x2+4x)+405x8+810x9-243x10=1+10x+40x2+80x3+80x4+32x5-15x2-120x3-3604-480x5-240x6+      90x4+720x7+540x5+1080x6-270x6-1080x8-1080x7+405x8+810x9-243x10=1+10x+25x2-40x3-190x4+92x5+570x6-360x7-675x8+810x9-243x10


(ix)
(x+1-1x)3=C03(x+1)3(1x)0-C13(x+1)2(1x)1+C23(x+1)1(1x)2-C33(x+1)0(1x)3
=(x+1)3-3(x+1)2×1x+3x+1x2-1x3=x3+1+3x+3x2-3x2+3+6xx+3x+1x2-1x3=x3+1+3x+3x2-3x-3x-6+3x+3x2-1x3=x3+3x2-5+3x2-1x3

(x)
(1-2x+3x2)3=C03(1-2x)3+C13(1-2x)2(3x2)+C23(1-2x)(3x2)2+C33(3x2)3=(1-2x)3+9x2(1-2x)2+27x4(1-2x)+27x6=1-8x3+12x2-6x+9x2(1+4x2-4x)+27x4-54x5+27x6=1-8x3+12x2-6x+9x2+36x4-36x3+27x4-54x5+27x6=1-6x+21x2-44x3+63x4-54x5+27x6

Page No 18.11:

Question 2:

Evaluate the following:
(i) x+1+x-16+x+1-x-16

(ii) x+x2-16+x-x2-16

(iii) 1+2 x5+1-2 x5

(iv) 2+16+2-16

(v) 3+25-3-25

(vi) 2+37+2-37

(vii) 3+15-3-15

(viii) 0.995+1.015

(ix) 3+26-3-26

(x) a2+a2-14+a2-a2-14

Answer:

(i)
(x+1+x-1)6+(x+1-x-1)6=2[C06 (x+1)6(x-1)0+C26 (x+1)4(x-1)2+C46 (x+1)2(x-1)4+C66 (x+1)0(x-1)6]=2[(x+1)3+15(x+1)2(x-1)+15(x+1)(x-1)2+(x-1)3=2[x3+1+3x+3x2+15(x2+2x+1)(x-1)+15(x+1)(x2+1-2x)+x3-1+3x-3x2]=2[2x3+6x+15x3-15x2+30x2-30x+15x-15+15x3+15x2-30x2-30x+15x+15]=2[32x3-24x]=16x[4x2-3]

(ii)
(x+x2-1)6+(x-x2-1)6=2[C06x6(x2-1)0+C26x4(x2-1)2+C46x2(x2-1)4+C66x0(x2-1)6]=2[x6+15x4(x2-1)+15x2(x2-1)2+(x2-1)3]=2[x6+15x6-15x4+15x2(x4-2x2+1)+(x6-1+3x2-3x4)]=2[x6+15x6-15x4+15x6-30x4+15x2+x6-1+3x2-3x4]=64x6-96x4+36x2-2

(iii)
(1+2x)5+(1-2x)5=2[C05(2x)0+C25(2x)2+C45(2x)4]=2[1+10×4x+5×16x2]=2[1+40x+80x2]

(iv)
(2+1)6+(2-1)6=2[C06(2)6+C26(2)4+C46(2)2+C66(2)0]=2[8+15×4+15×2+1)=2×99 =198

(v)
(3+2)5-(3-2)5=2C15×34×(2)1+C35×32×(2)3+C55×30×(2)5

=2[5×81×2+10×9×22+42]=22(405+180+4)=11782

(vi)
(2+3)7+(2-3)7=2[C07×27×(3)0+C27×25×(3)2+C47×23×(3)4+C67×21×(3)6]=2[128+21×32×3+35×8×9+7×2×27]=2[128+2016+2520+378]=2×5042=10084

(vii)
(3+1)5-(3-1)5=2[C15×(3)4+C35×(3)2+C55×(3)0]=2[5×9+10×3+1]=2×76=152

(viii)
(0.99)5+(1.01)5=(1-0.01)5+(1+0.01)5=2[C05(0.01)0+C25(0.01)2+C45(0.01)4]=2[1+10×0.0001+5×0.00000001]=2×1.00100005=2.0020001

(ix)
(3+2)6-(3-2)6=2[C16(3)5(2)1+C36(3)3(2)3+C56(3)1(2)5]
=2[6×93×2+20×33×22+6×3×42]=2[6(54+120+24)]=3966

(x)
a2+a2-14+a2-a2-14=2[C04(a2)4(a2-1)0+C24(a2)2(a2-1)2+C44(a2)0(a2-1)4]=2[a8+6a4(a2-1)+(a2-1)2]=2[a8+6a6-6a4+a4+1-2a2]=2a8+12a6-10a4-4a2+2

Page No 18.11:

Question 3:

Find a+b4-a-b4. Hence, evaluate 3+24-3-24.

Answer:

The expression (a+b)4-(a-b)4 can be written as

(a+b)4-(a-b)4=2[C14a3b1+C34a1b3]                                =2[4a3b+4ab3]                               =8(a3b+ab3)

Putting a=3 and b =2, we get: (3+2)4-(3-2)4=8[(3)3×2+3×(2)3]                                               =8(36+26)                                               =406

 (3+2)4-(3-2)4=406                                               

Page No 18.11:

Question 4:

Find x+16+x-16. Hence, or otherwise evaluate 2+16+2-16.

Answer:

The expression x+16+x-16 can be written as
(x+1)6+(x-1)6=2[C06x6+C26x4+C46x2+C66x0]=2[x6+15x4+15x2+1]

By taking x=2, we get:
(2+1)6+(2-1)6=2[(2)6+15(2)4+15(2)2+1]
                                 =2[8+15×4+15×2+1]=2×(8+60+30+1)=198



Page No 18.12:

Question 5:

Using binomial theorem evaluate each of the following:
(i) (96)3
(ii) (102)5
(iii) (101)4
(iv) (98)5

Answer:

(i) (96)3
=(100-4)3=C03×1003×40-C13×1002×41+C23×1001×42-C33×1000×43=1000000-120000+4800-64=884736

(ii) (102)5
=(100+2)5=C05×1005×20+C15×1004×21+C25×1003×22+C35×1002×23+C45×1001×24+C55×1000×25=10000000000+1000000000+40000000+800000+8000+32=11040808032

(iii) (101)4
=(100+1)4=C04×1004+C14×1003+C24×1002+C34×1001+C44×1000=100000000+4000000+60000+400+1=104060401

(iv) (98)5
(100-2)5=C05×1005×20+-5C1×1004×21+C25×1003×22-C35×1002×23+C45×1001×24-C55×1000×25=10000000000-1000000000+40000000-800000+8000-32=9039207968

Page No 18.12:

Question 6:

Using binomial theorem, prove that 23n-7n-1 is divisible by 49, where n  N.

Answer:

23n-7n-1=8n-7n-1                ...(1)

Now,8n=(1+7)n    =C0n+C1n×71+C2n×72+C3n×73+C4n×74+...+Cnn×7n8n=1+7n+49[C2n+C3n×71+C4n×72+...+Cnn×7n-2]8n-1-7n=49×An integerNow, 8n-1-7n is divisible by 49Or,23n-1-7n is divisible by 49     From (1)

Page No 18.12:

Question 7:

Using binomial theorem, prove that 32n+2-8n-9 is divisible by 64, n  N.

Answer:

32n+2-8n-9=9n+1-8n-9       ...1
Consider
 9n+1=1+8n+19n+1 =C0n+1×80+C1n+1×81+C2n+1×82+C3n+1×83+...+Cn+1n+1×8n+1
9n+1=1+8(n+1)+[C2n+1×82+C3n+1×83+...+Cn+1n+1×8n+1]9n+1-8n-9=64(C2n+1+C3n+1×81+...+Cn+1n+1×8n-1]9n+1-8n-9=64×An integer9n+1-8n-9 is divisible by 64Or,32n+2-8n-9 is divisible by 64    From (1)Hence proved.

Page No 18.12:

Question 8:

If n is a positive integer, prove that 33n-26n-1 is divisible by 676.

Answer:

33n-26n-1=27n-26n-1       ...1
Now, we have:27n=(1+26)nOn expanding, we get(1+26)n=C0n×260+C1n×261+C2n×262+C3n×263+C4n×264+...Cnn×26n27n=1+26n+262[C2n+C3n×261+C4n×262+...Cnn×26n-2]27n-26n-1=676×an integer27n-26n-1 is divisible by 676Or,33n-26n-1 is divisible by 676   From (1)

Page No 18.12:

Question 9:

Using binomial theorem, indicate which is larger (1.1)10000 or 1000.

Answer:

We have:   
(1.1)10000  
=(1+0.1)10000=C010000×0.10+C110000×(0.1)1+C210000×(0.1)2+...C1000010000×(0.1)10000=1+10000×0.1+other positive terms=1+10000+other positive terms=10001+other positive terms10001>1000(1.1)10000>1000

Page No 18.12:

Question 10:

Using binomial theorem determine which number is larger (1.2)4000 or 800?

Answer:

We have:
(1.2)4000=(1+0.2)4000=C04000+C14000×(0.2)1+C24000×(0.2)2+...C40004000×(0.2)4000

              =1+4000×0.2+other positive terms=1+800+other positive terms=801+other positive terms801 >800

Hence, (1.2)4000 is greater than 800

Page No 18.12:

Question 11:

Find the value of (1.01)10 + (1 − 0.01)10 correct to 7 places of decimal.

Answer:

(1.01)10+(1-0.01)10=(1+0.01)10+(1-0.01)10=2[C010×(0.01)0+C210×(0.01)2+C410×(0.01)4+C610×(0.01)6+C810×(0.01)8+C1010×(0.01)10]=21+45×0.0001+210×0.00000001+... =21+0.0045+0.00000210+...=2.0090042+...

Hence, the value of (1.01)10 + (1 − 0.01)10 correct to 7 places of the decimal is 2.0090042

Page No 18.12:

Question 12:

Show that 24n+4-15n-16, where n ∈  is divisible by 225.

Answer:

We have,
24n+4-15n-16=24n+1-15n-16                        =16n+1-15n-16                        =1+15n+1-15n-16                        =n+1C0 150+n+1C1 151+n+1C2 152+...+n+1Cn+1 15n+1-15n-16                        =1+(n+1)15+n+1C2 152+...+n+1Cn+1 15n+1-15n-16                        =1+15n+15+n+1C2 152+...+n+1Cn+1 15n+1-15n-16                        =n+1C2 152+...+n+1Cn+1 15n+1                        =152n+1C2+...+n+1Cn+1 15n-1                        =225n+1C2+...+n+1Cn+1 15n-1

Thus, â€‹ 24n+4-15n-16, where n ∈  is divisible by 225.



Page No 18.37:

Question 1:

Find the 11th term from the beginning and the 11th term from the end in the expansion of 2x-1x225.

Answer:

Given:
2x-1x225
Clearly, the given expression contains 26 terms.

So, the 11th term from the end is the (26 − 11 + 1)th term from the beginning. In other words, the 11th term from the end is the 16th term from the beginning.

Thus, we have:
T16=T15+1=C1525(2x)25-15-1x215                    =C1525210x10-1x30=-C1525210x20
Now, we will find the 11th term from the beginning.

T11=T10+1      =C1025(2x)25-10-1x210      =C1025215x151x20      =C1025215x5

Page No 18.37:

Question 2:

Find the 7th term in the expansion of 3x2-1x310.

Answer:

We need to find the 7th term of the given expression.
Let it be T7
Now, we have
T7=T6+1
    =C610(3x2)10-6-1x36=C61034x81x18=10×9×8×7×814×3×2×x10=17010x10

Thus, the 7th term of the given expression is 17010x10

Page No 18.37:

Question 3:

Find the 5th term from the end in the expansion of 3x-1x210

Answer:

Given:
3x-1x210
Clearly, the expression has 6 terms.
The 5th term from the end is the (11 − 5 + 1)th, i.e., 7th, term from the beginning.
Thus, we have:

T7=T6+1=C610(3x)10-6-1x26=C61034x41x12=10×9×8×7×814×3×2×1×x8=17010x8

Page No 18.37:

Question 4:

Find the 8th term in the expansion of x3/2 y1/2- x1/2 y3/210.

Answer:

We need to find the 8th term in the given expression.
T8=T7+1

T8=C710(x3/2y1/2)10-7(-x1/2y3/2)7    =-10×9×83×2x9/2y3/2x7/2y21/2        =-120x8y12

Page No 18.37:

Question 5:

Find the 7th term in the expansion of 4x5+52x8.

Answer:

We need to find the 7th term in the given expression.

T7=T6+1

T7=T6+1    =C684x58-652x6   =8×7×4×4×125×1252×1×25×64 x2 1x6  =4375x4

Page No 18.37:

Question 6:

Find the 4th term from the beginning and 4th term from the end in the expansion of x+2x9.

Answer:

Let Tr+1 be the 4th term from the end.
Then, Tr+1 is (10 − 4 + 1)th, i.e., 7th, term from the beginning.

  T7=T6+1          =C69 x9-62x6         =9×8×73×2x364x6         =5376x3

4th term from the beginning = T4=T3+1
T4=C39 x9-3 2x3    =9×8×73×2x68x3    =672 x3

Page No 18.37:

Question 7:

Find the 4th term from the end in the expansion of 4x5-52x8.

Answer:

Let Tr+1 be the4th term from the end of the given expression.
Then,
Tr+1 is (10 − 4 + 1)th term, i.e., 7th term, from the beginning.
Thus, we have:
T7=T6+1    =C69 4x59-652x6   =9×8×73×264125x3125×12564x6  =10500x3

Page No 18.37:

Question 8:

Find the 7th term from the end in the expansion of 2x2-32x8

Answer:

Let Tr+1 be the 7th term from the end in the given expression.
Then, we have:
Tr+1 = (9 − 7 + 1) =  3rd term from the beginning
Now,
T3=T2+1    =C28 (2x2)8-2 -32x2    =8×72×164x1294x2   =4032 x10

Page No 18.37:

Question 9:

Find the coefficient of:
(i) x10 in the expansion of 2x2-1x20

(ii) x7 in the expansion of x-1x240

(iii) x-15 in the expansion of 3x2-a3x310

(iv) x9 in the expansion of x2-13x9

(v) xm in the expansion of x+1xn

(vi) x in the expansion of 1-2x3+3x5 1+1x8.

(vii) a5b7 in the expansion of a-2b12.

(viii) x in the expansion of 1-3x+7x2 1-x16.

Answer:

(i) Suppose x10 occurs in the (+ 1)th term in the given expression.

Then, we have:

Tr+1=Crn xn-r ar  

Here,
Tr+1=Cr20(2x2)20-r -1xr         =(-1)r Cr20220-r x40-2r-rFor this term to contain x10, we must have:40-3r =103r=30r=10 Coefficient of x10 = (-1)10  C1020220-10 =C1020210 

(ii) Suppose x7 occurs at the (+ 1) th term in the given expression.

Then, we have:
Tr+1=Cr40 x40-r-1x2r         =(-1)r  Cr40 x40-r-2rFor this term to contain x7, we must have:40-3r=73r=40-7=33r=11Coefficient of x7 = (-1)11  C1140=-C1140

(iii)  Suppose x−15 occurs at the (+ 1)th term in the given expression.
Then, we have:
 Tr+1=Cr10 (3x2)10-r-a3x3r
Tr+1=(-1)r Cr10 310-r-rx20-2r-3rar
For this term to contain x-15 , we must have:20-5r =-155r=20+15r=7Coefficient of x-15 = (-1)7  C710 310-14 a7=-10×9×83×2×9×9a7=-4027a7

(iv) Suppose x9 occurs at the (+ 1)th term in the above expression.

Then, we have:

Tr+1=Cr9 (x2)9-r -13xr         =(-1)r Cr9 x18-2r-r 13rFor this term to contain x9, we must have:18-3r=93r=9r=3Coefficient of x9 =(-1)3 C39 133=-9×8×72×9×9=-289

(v)
Suppose xm occurs at the (+ 1)th term in the given expression.

Then, we have:

Tr+1=Crn xn-r 1xr=Crn xn-2rFor this term to contain xm, we must have:n-2r =mr=(n-m)/2Coefficient of xm = C(n-m)/2n=n!n-m2! n+m2!

(vi) Suppose x occurs at the (+ 1)th term in the given expression.

Then, we have:

(1-2x3+3x5)1+1x8=1-2x3+3x5C08+C18 1x+C28 1x2+C38 1x3+C48 1x4+C58 1x5+C68 1x6+C78 1x7+C88 1x8 x occurs in the above expresssion at -2x3.C28 1x2 +3x5.C48 1x4.Coefficient of x =-28!2! 6!+38!4! 4!=-56+210= 154
 
(vii)
Suppose a5 b7 occurs at the (r + 1)th term in the given expression.

Then, we have:

Tr+1=Cr12 a12-r (-2b)r=(-1)r Cr12 a12-r br 2rFor this term to contain a5 b7, we must have:12-r =5  r=7 Required coefficient = (-1)7 C712 27=-12×11×10×9×8×1285×4×3×2=-101376

(viii) Suppose x occurs at the (+ 1)th term in the given expression.

Then, we have:

1-3x+7x2 1-x16=1-3x+7x2C016+C116 -x+C216 -x2+C316 -x3+C416 -x4+16C5 -x5+C616 -x6+C716 -x7+C816 -x8+C916 -x9+C1016 -x10+C1116 -x11+C1216 -x12+C1316 -x13+C1416 -x14+C1516 -x15+C1616 -x16 x occurs in the above expresssion at 16C1 -x -3xC016.Coefficient of x =-16!1! 15!-316!0! 16!=-16-3= -19



Page No 18.38:

Question 10:

Which term in the expansion of xy1/3+yx1/31/221 contains x and y to one and the same power?

Answer:

Suppose Tr+1th term in the given expression contains x and y to one and the same power.
Then,

Tr+1 th term isCr21xy1/321-r yx1/31/2r=Cr21 x(21-r)/3xr/6yr/2y(21-r)/6=Cr21 x7-r/2y2r/3-7/2Now, if x and y have the same power, then7-r2=2r3-722r3+r2=7+727r6=212r=9Hence, the required term is the 10th term

Page No 18.38:

Question 11:

Does the expansion of 2x2-1x contain any term involving x9?

Answer:

Suppose x9 occurs in the given expression at the (r + 1)th term.
Then, we have:
Tr+1=Cr20 (2x2)20-r -1xr=(-1)r  Cr20220-r x40-2r-rFor this term to contain x9, we must have40-3r=93r=31r=313 It is not possible, as r is not an integer.

Hence, there is no term with x9 in the given expression.

Page No 18.38:

Question 12:

Show that the expansion of x2+1x12 does not contain any term involving x−1.

Answer:

Suppose x1 occurs at the (r + 1)th term in the given expression.
Then,
Tr+1=Cr12 (x2)12-r 1xr=Cr12 x24-2r-rFor this term to contain x-1, we must have24-3r=-13r=25r=253It is not possible, as r is not  an integer.

Hence, the expansion of x2+1x12 does not contain any term involving x−1.

Page No 18.38:

Question 13:

Find the middle term in the expansion of:
(i) 23x-32x20

(ii) ax+bx12

(iii) x2-2x10

(iv) xa-ax10

Answer:

(i) Here,
n = 20  (Even number) 
Therefore, the middle term is the n2+1th term, i.e., the 11th term.
Now,T11=T10+1=C1020  23x20-10 32x10=C1020  210310×310210x10-10=C1020

(ii) Here,
n = 12 (Even number)
Therefore, the middle term is the n2+1th  i.e. 7th term
Now,T7=T6+1=C612 ax12-6 (bx)6=C612 a6 b6 =12×11×10×9×8×76×5×4×3×2a6 b6=924 a6b6

(iii) Here,
n = 10    (Even number)
Therefore, the middle term is the n2+1th  i.e. 6th term
Now,T6=T5+1=C510 (x2)10-5 -2x5=-10×9×8×7×65×4×3×2×32x5=-8064 x5

(iv) Here,
n = 10 (Even number)
Therefore, the middle term is the n2+1th  i.e. 6th term
Now,T6=T5+1         =C510 xa10-5 -ax5         =-10×9×8×7×65×4×3×2=-252

Page No 18.38:

Question 14:

Find the middle terms in the expansion of:
(i) 3x-x369

(ii) 2x2-1x7

(iii) 3x-2x215

(iv) x4-1x311

Answer:

(i) Here, n, i.e. 9, is an odd number.
Thus, the middle terms are n+12th and n+12+1th, i.e. 5th and 6th
Now,T5=T4+1=C49 (3x)9-4 -x364=9×8×7×64×3×2×27×9×136×36x17=1898x17and,T6=T5+1=C59(3x)9-5 -x365=-9×8×7×64×3×2×81×1216×36x19=-2116x19

(ii) Here, n, i.e., 7, is an odd number.

Thus, the middle terms are 7+12th and 7+12+1th i.e. 4th and 5thNow,T4=T3+1=C37 (2x2)7-3 -1x3=-7×6×53×2×16 x8×1x3=-560 x5And,T5=T4+1=C47 (2x2)7-4 -1x4=35×8 ×x6×1x4=280 x2

(iii)
Given:n, i.e.15  is an odd number.Thus, the middle terms are 15+12th and 15+12+1th i.e. 8th and 9th.Now,T8= T7+1    =C715 (3x)15-7 -2x27    =-15×14×13×12×11×10×97×6×5×4×3×2×38×27 x8-14    =-6435×38×27x6And,T9=T8+1    =C815 (3x)15-8 -2x28    =15×14×13×12×11×10×97×6×5×4×3×2×37×28 ×x7-16    =6435×37×28x9

(iv)
Here, n, i.e., 11, is an odd number.Thus, the middle terms are 11+12th and 11+12+1th i.e. 6th and 7th.Now,T6=T5+1    =C511 (x4)11-5 -1x35    =-11×10×9×8×75×4×3×2×x24-15    =-462 x9And,T7=T6+1     =C611 (x4)11-6 -1x36     =11×10×9×8×75×4×3×2x20-18     =462 x2

Page No 18.38:

Question 15:

Find the middle terms(s) in the expansion of:
(i) x-1x10

(ii) 1-2x+x2n

(iii) 1+3x+3x2+x32n

(iv) 2x-x249

(v) x-1x2n+1

(vi) x3+9y10

(vii) 3-x367

(viii) 2ax-bx212

(ix) px+xp9

(x) xa-ax10

Answer:

(i)
x-1x10Here, n is an even number.  Middle term = 102+1th= 6th termNow, we haveT6=T5+1=C510 x10-5 -1x5=-10×9×8×7×65×4×3×2=-252

(ii)
(1-2x+x2)n=(1-x)2nn is an even number. Middle term = 2n2+1th=(n+1)th termNow, we haveTn+1=Cn2n (-1)n (x)n=(2n)!(n!)2(-1)n xn

(iii)
(1+3x+3x2+x3)2n=(1+x)6nHere, n  is an even number. Middle term = 6n2+1 th=(3n+1)th termNow, we haveT3n+1=C3n6n x3n=(6n)!(3n!)2x3n

(iv)
2x-x249Here, n is an odd number.Therefore, the middle terms are n+12th and n+12+1th, i.e. 5th and 6th terms.Now, we haveT5=T4+1=C49 (2x)9-4 -x244=9×8×7×64×3×2×25 144x5+8=634x13And,T6=T5+1=C59 (2x)9-5 -x245=-9×8×7×64×3×2×24 145x4+10=-6332x14

(v)
x-1x2n+1Here, 2n+1  is an odd number.Therefore, the middle terms are 2n+1+12th and2n+1+12+1th i.e. (n+1)th and (n+2)th terms.Now, we have:Tn+1=Cn2n+1 x2n+1-n  × (-1)nxn=(-1)n Cn2n+1 xAnd,Tn+2=Tn+1+1=Cn+12n+1 x2n+1-n-1   (-1)n+1xn+1=(-1)n+1 Cn+12n+1 ×1x

(vi)
x3+9y10Here, n is an even number.Therefore, the middle term is 102+1th, i.e., 6th term.Now, we haveT6=T5+1=C510 x310-5 (9y)5=10×9×8×7×65×4×3×2×135×95×x5 y5=61236 x5 y5

(vii)
3-x367Here, n is an odd number.Therefore, the middle terms are 7+12th and 7+12+1th, i.e., 4th and 5th terms.Now, we haveT4=T3+1=C37 37-3 -x363=-1058x9And,T5=T4+1=C49 39-4 -x364=7×6×53×2×35×164 x12=3548x12

(viii)
2ax-bx212Here, n is an even number.  Middle term = 122+1th= 7th termNow, we haveT7=T6+1=C612 2ax12-6 -bx26=12×11×10×9×8×76×5×4×3×2×1×2abx6=59136 a6b6x6

(ix)
px+xp9Here, n is an odd number.Therefore, the middle terms are 9+12th and 9+12+1th, i.e., 5th and 6th terms.Now, we haveT5=T4+1=C49 px9-4 xp4=9×8×7×64×3×2×1×px=126 pxAnd,T6=T5+1=C59 px9-5 xp5=9×8×7×64×3×2×1×xp=126 xp

(x)
xa-ax10Here, n is an even number.  Middle term = 102+1th= 6th termNow, we haveT6=T5+1=C510 xa10-5 -ax5=-10×9×8×7×65×4×3×2×1=-252



Page No 18.39:

Question 16:

Find the term independent of x in the expansion of the following expressions:
(i) 32x2-13x9

(ii) 2x+13x29

(iii) 2x2-3x325

(iv) 3x-2x215

(v) x3+32x210

(vi) x-1x23n

(vii) 12x1/3+x-1/58

(viii) 1+x+2x332x2-33x9

(ix) x3+12 x318, x > 0

(x) 32x2-13x6

Answer:

(i) Suppose the (r + 1)th term in the given expression is independent of x.
Now,
32x2-13x9Tr+1=Cr9 32x29-r -13xr= (-1)r Cr9 .39-2r29-r× x18-2r-rFor this term to be independent of x, we must have18-3r=03r=18r=6Hence, the required term is the 7th term.Now, we haveC69×39-1229-6=9×8×73×2×3-3×2-3=718

(ii) Suppose the (r + 1)th term in the given expression is independent of x.
Now,
2x+13x29Tr+1=Cr9 (2x)9-r13x2r=Cr9.29-r3r x9-r-2rFor this term to be independent of x, we must have9-3r=0r=3Hence, the required term is the 4th term.Now, we haveC39 2633=C39×6427

(iii) Suppose the (r + 1)th term in the given expression is independent of x.
Now,
2x2-3x325Tr+1=Cr25 (2x2)25-r -3x3r=(-1)r  Cr25 ×225-r×3r x50-2r-3rFor this term to be independent of x, we must have:50-5r=0r=10Therefore, the required term is the 11th term.Now, we have(-1)10  C1025 ×225-10×310=C1025 (215× 310)

(iv)  Suppose the (r + 1)th term in the given expression is independent of x.
Now,
3x-2x215Tr+1=Cr15 (3x)15-r -2x2r= (-1)r  Cr15 ×315-r × 2r x15-r-2rFor this term to be independent of x, we must have15-3r=0r=5Hence, the required term is the 6th term.Now, we have:(-1)5  C515 .315-5 . 25=-3003 ×310×25

(v) Suppose the (r + 1)th term in the given expression is independent of x.
Now,
x3+32x210Tr+1=Cr10 x310-r 32x2r=Cr10 .3r-10-r22r x10-r2-2rFor this term to be independent of x, we must have10-r2-2r=010-5r=0r=2Hence, the required term is the 3rd term.Now, we haveC210 ×32-10-2222=10×92×4×9=54

(vi)  Suppose the (r + 1)th term in the given expression is independent of x.
Now,
x-1x23nTr+1=Cr3n x3n-r -1x2r=(-1)r Cr3n  x3n-r-2rFor this term to be independent of x, we must have3n-3r=0r = nHence, the required term is the (n+1)th term.Now, we have(-1)n Cn3n


(vii)  Suppose the (r + 1)th term in the given expression is independent of x.
Now,
12x1/3 +x-1/58Tr+1=Cr8 12x1/38-r (x-1/5)r=Cr8. 128-r x8-r3-r5For this term to be independent of x, we must have8-r3-r5=040-5r-3r=08r=40r=5Hence, the required term is the 6th term.Now, we have:C58× 128-5=8×7×63×2×8=7

(ix) Suppose the (r + 1)th term in the given expression is independent of x.
Now,
x3+12x318Tr+1=Cr18 (x1/3)18-r 12 x1/3r=Cr18×12r x18-r3-r3For this term to be independent of r, we must have18-r3-r3=018-2r=0r=9The term is C918×129

(x) Suppose the (r + 1)th term in the given expression is independent of x.
Now,
32x2-13x6Tr+1=Cr6 32x26-r -13xr=-1r Cr6 × 36-r-r26-r x12-2r-rFor this term to be independent of x, we must have12-3r=0r=4Hence, the required term is the 4th term.C46 × 36-4-426-4=6×52×1×4×9=512

Page No 18.39:

Question 17:

If the coefficients of 2r + 4th and r-2th terms in the expansion of 1+x18 are equal, find r.

Answer:

Given:(1+x)18We know that the coefficient of the rth term in the expansion of (1+x)n  is Cr-1nTherefore, the coefficients of the (2r+4)th and (r-2)th terms in the given expansion are C2r+4-1 18and Cr-2-118For these coefficients to be equal, we must haveC2r+3 18= Cr-3182r+3=r-3  or, 2r+3+r-3=18      [ Crn=Csn r=s or r+s=n]r=-6   or,  r=6Neglecting negative value We getr=6

Page No 18.39:

Question 18:

If the coefficients of (2r + 1)th term and (r + 2)th term in the expansion of (1 + x)43 are equal, find r.

Answer:

Given :- (1+x)43We know that the coefficient of the rth term in the expansion of (1+x)n is Cr-1nTherefore, the coefficients of the (2r+1)th and (r+2)th terms in the given expression are C2r+1-143 and Cr+2-143For these coefficients to be equal, we must have:2r=r+1   or,  2r+r+1=43      [ Crn=Csn r=s or r+s=n] r=14    for r=1 it gives the same term

Page No 18.39:

Question 19:

Prove that the coefficient of (r + 1)th term in the expansion of (1 + x)n + 1 is equal to the sum of the coefficients of rth and (r + 1)th terms in the expansion of (1 + x)n.

Answer:

Coefficient of the (r+1)th term in (1+x)n+1 is Crn+1Sum of the coefficients of the rth and (r+1)th terms in (1+x)n=Cr-1n +Crn                                                                                           =Crn+1           Cr+1n +Crn=Cr+1n+1  Hence proved.

Page No 18.39:

Question 20:

Prove that the term independent of x in the expansion of x+1x2n is 1·3·5 ... 2n-1n!. 2n.

Answer:

Given:x+1x2nSuppose the term independent of x is the (r+1) th term.Tr+1=Cr2n x2n-r 1xr         =Cr2n x2n-2rFor this term to be independent of x, we must have:2n-2r=0n=rRequired coefficient = Cn2n                                        =(2n)!(n!)2                                         =1·3·5...2n-32n-12·4·6...2n-22n(n!)2                                        =1·3·5...2n-32n-12nn!

Page No 18.39:

Question 21:

The coefficients of 5th, 6th and 7th terms in the expansion of (1 + x)n are in A.P., find n.

Answer:

Coefficients of the 5th, 6th and 7th terms in the given expansion are C4n , C5n and C6nThese coefficients are in AP. Thus, we have2 C5n =C4n+C6nOn dividing both sides by C5n, we get:2=C4nC5n+C6nC5n2=5n-4+n-5612n-48=30+n2-4n-5n+20n2-21n+98=0(n-14)(n-7)=0n=7 or 14

Page No 18.39:

Question 22:

If the coefficients of 2nd, 3rd and 4th terms in the expansion of (1 + x)2n are in A.P., show that 2n2-9n+7=0.

Answer:

Given:(1+x)2nThus, we have:T2=T1+1     =C12n x1T3=T2+1    =C22n x2T4=T3+1    =C32n x3We have coefficients of the 2nd, 3rd and 4th terms in AP.2C22n =C12n+C32n 2=C12nC22n+C32nC22n                 2=22n-1+2n-2312n-6=6+4n2-4n-2n+24n2-18n+14=02n2-9n+7=0

Hence proved.

Page No 18.39:

Question 23:

If the coefficients of 2nd, 3rd and 4th terms in the expansion of (1 + x)n are in A.P., then find the value of n.

Answer:

Coefficients of the 2nd, 3rd  and 4th terms in the given expansion are:
C1n, C2n and C3nWe have:2×C2n=C1n+C3nDividing both sides by C2n, we get:2=C1nC2n+C3nC2n2=2n-1+n-236n-6=6+n2+2-3nn2-9n+14=0 n=7  n2 as 2>3 in the 4th term

Page No 18.39:

Question 24:

If in the expansion of (1 + x)n, the coefficients of pth and qth terms are equal, prove that p + q = n + 2, where pq.

Answer:

Coefficients of the pth and qth terms are Cp-1n and Cq-1n respectively.Thus, we have:Cp-1n= Cq-1np-1 =q-1 or, p-1+q-1=n             [ Crn=nCsr=s or, r+s=n]p=q or, p+q=n+2

If pq, then p+q=n+2
Hence proved

Page No 18.39:

Question 25:

Find a, if the coefficients of x2 and x3 in the expansion of (3 + ax)9 are equal.

Answer:

(3+ax)9=C09 .39. (ax)0 +C19 .38. (ax)1+C29 .37. (ax)2+C39 .36. (ax)3+...

We have Coefficient of x2Coefficient of x3

C29 ×37 a2 =C39 ×36 a3a=C29C39×3       =9! ×3! ×6!×32! × 7! ×9!       =97

Page No 18.39:

Question 26:

Find the coefficient of a4 in the product (1 + 2a)4 (2 − a)5 using binomial theorem.

Answer:

(1+2a)4(2-a)5=[C04 (2a)0+C14 (2a)1+C24 (2a)2+C34 (2a)3+C44 (2a)4]×            [C05 (2)5 (-a)0 +C15 (2)4 (-a)1+C25 (2)3 (-a)2+C35 (2)2 (-a)3+C45 (2)1 (-a)4+C55 (2)0 (-a)5]=[1+8a+24a2+32a3+16a4]×[32-80a+80a2-40a3+10a4-a5]Coefficient of a4=10-320+1920-2560+512=-438



Page No 18.40:

Question 27:

In the expansion of (1 + x)n the binomial coefficients of three consecutive terms are respectively 220, 495 and 792, find the value of n.

Answer:

Suppose the three consecutive terms are Tr-1, Tr and Tr+1.Coefficients of these terms are Cr-2n, Cr-1n and Crn, respectively.These coefficients are equal to 220, 495 and 792. Cr-2n Cr-1n=220495r-1n-r+2=499r-9=4n-4r+84n+17=13r      ...1Also,CrnCr-1n=792495n-r+1r=855n-5r+5=8r5n+5=13r 5n+5=4n+17         From Eqn1n=12 

Page No 18.40:

Question 28:

If in the expansion of (1 + x)n, the coefficients of three consecutive terms are 56, 70 and 56, then find n and the position of the terms of these coefficients.

Answer:

Suppose rth, (r+1) thand (r+2)th terms are the three consecutive terms.Their respective coefficients are Cr-1n,  Crn and Cr+1n.We have: Cr-1n=Cr+1n=56r-1+r+1=n       [If Crn=Csnr=s or r+s =n]2r=nr=n2Now,Cn2n =70 and Cn2-1n=56Cn2-1n Cn2n=5670n2n2+1=8105n=4n+8n=8So, r=n2=4Thus, the required terms are 4th, 5th and 6th.

Page No 18.40:

Question 29:

If 3rd, 4th 5th and 6th terms in the expansion of (x + a)n be respectively a, b, c and d, prove that b2-acc2-bd=5a3c.

Answer:

  We have:(x+a)nThe 3rd, 4th, 5th and 6th terms are C2nxn-2 a2,  C3nxn-3 a3, C4nxn-4 a4 and C5nxn-5 a5, respectively.Now,C2nxn-2 a2=aC3nxn-3 a3 =bC4nxn-4 a4 =cC5nxn-5 a5 =dLHS =b2-acc2-bd

Page No 18.40:

Question 30:

If a, b, c and d in any binomial expansion be the 6th, 7th, 8th and 9th terms respectively, then prove that b2-acc2-bd=4a3c.

Answer:

Suppose the binomial expression is (1+x)n.Then, the 6th, 7th, 8th and 9th terms are C5n x5, C6n x6,  C7n x7 and C8n x8, respectively.Now, we have: C6 x6n C5n x5=ba,  C8 x8n C7n x7=dc and  C7 x7n C6n x6=cbn-56=ba and n-67=cbn-67n-56=cbba6n-367n-35=ca

Page No 18.40:

Question 31:

If the coefficients of three consecutive terms in the expansion of (1 + x)n be 76, 95 and 76, find n.

Answer:

Suppose r, r+1 and r+2 are three consecutive terms in the given expansion.The coefficients of these terms are Cr-1n, Crn and Cr+1n.According to the question,Cr-1n=76Crn=95Cr+1n=76Cr-1n=Cr+1nr-1+r+1=n             [If Crn=Csnr=s or r+s =n]r=n2CrnCr-1n=9576n-r+1r=9576n2 +1n2=957638n+76=95n2       19n2=76n=8

Page No 18.40:

Question 32:

If the 6th, 7th and 8th terms in the expansion of (x + a)n are respectively 112, 7 and 1/4, find x, a, n.

Answer:

The 6th, 7th and 8th terms in the expansion of (x+a)n are C5n xn-5 a5, C6n xn-6 a6 and C7n xn-7 a7.
According to the question,
C5n xn-5 a5=112C6n xn-6 a6=7C7n xn-7 a7=14Now,C6 nxn-6 a6C5n xn-5 a5=7112n-6+16x-1 a=116ax=38n-40                    ...1Also,C7 nxn-7 a7C6n xn-6 a6=1/47n-7+17x-1a=128ax=14n-24                        ...2From 1 and 2, we get: 38n-40=14n-2432n-10=1n-6n=8


Putting in eqn1 we geta=xNow, C58 x8-5 x85=11256x885=112x8=48x=4By putting the value of x and n in 1 we geta=12

a=3 and x=2

Page No 18.40:

Question 33:

If the 2nd, 3rd and 4th terms in the expansion of (x + a)n are 240, 720 and 1080 respectively, find x, a, n.

Answer:

In the expansion of x+an, the 2nd, 3rd and 4th terms are C1n xn-1 a1, C2n xn-2 a2 and C3n xn-3 a3, respectively.According to the question,C1n xn-1 a1=240        C2n xn-2 a2=720C3n xn-3 a3=1080C2n xn-2 a2C1n xn-1 a1=720240n-12xa=3ax=6n-1    ...1Also,C3n xn-3 a3C2n xn-2 a2=1080720n-23xa=32ax=92n-4    ...2Using 1 and 2 we get6n-1=92n-4n=5Putting in eqn1 we get2a=3xNow, C15 x5-1 32x=24015x5=480x5=32x=2By putting the value of x and n in 1 we geta=3

Page No 18.40:

Question 34:

Find a, b and n in the expansion of (a + b)n, if the first three terms in the expansion are 729, 7290 and 30375 respectively.

Answer:

We have:T1=729, T2=7290 and T3=30375Now,C0n an b0=729an=729an=36C1n an-1 b1=7290C2n an-2 b2=30375Also,C2n an-2 b2C1n an-1 b1=303757290n-12×ba=256              ...(i)(n-1)ba=253And,C1n an-1 b1C0n an b0=7290729nba=101          ...(ii)On dividing (ii) by (i), we getnba(n-1)ba=10×325nn-1=65n = 6Since, a6=36 Hence, a=3Now,  nba=10b=5

Page No 18.40:

Question 35:

If the term free from x in the expansion of x-kx210 is 405, find the value of k.

Answer:

Let (r + 1)th term, in the expansion of x-kx210, be free from x and be equal to Tr + 1. Then,
Tr+1=10Crx10-r-kx2r=10Cr x5-5r2-kr        ....(1)

If Tr + 1 is independent of x, then
5-5r2=0r=2

Putting r = 2 in (1), we obtain

T3=10C2 -k2=45k2

But it is given that the value of the term free from x is 405.
45k2=405k2=9k=±3

Hence, the value of k is ±3.

Page No 18.40:

Question 36:

Find the sixth term in the expansion y12+x13n, if the binomial coefficient of the third term from the end is 45.

Answer:

In the binomial expansion of y12+x13n, there are (n + 1) terms.

The third term from the end in the expansion of y12+x13n, is the third term from the beginning in the expansion of x13+y12n.

∴ The binomial coefficient of the third term from the end = nC2

If is given that the binomial coefficient of the third term from the end is 45.

 nC2=45nn-12=45n2-n-90=0n-10n+9=0n=10       n cannot be negative

Let Tbe the sixth term in the binomial expansion of y12+x13n. Then
T6=nC5y12n-5x135=10C5 y52x53=252 y52x53

Hence, the sixth term in the expansion of y12+x13n, is 252 y52x53.

Page No 18.40:

Question 37:

If p is a real number and if the middle term in the expansion of p2+28 is 1120, find p.

Answer:

In the binomial expansion of p2+28, we observe that 82+1th i.e., 5th term is the middle term.

It is given that the middle term is 1120.

 T5=11208C4 p28-424=1120p4=16p=±2

Hence, the real values of p is ±2.

Page No 18.40:

Question 38:

Find n in the binomial 23+133n, if the ratio of 7th term from the beginning to the 7th term from the end is 16.

Answer:

In the binomail expansion of 23+133nn+1-7+1th i.e., (n − 5)th term from the beginning is the 7th term from the end.

Now,
T7=nC623n-61336=nC6×2n3-2×132And,Tn-5=nCn-6236133n-6=nC6×22×13n3-2

It is given that,
T7Tn-5=16nC6×2n3-2×132nC6×22×13n3-2=162n3-2-2×3n3-2-2=16164-n3=164-n3=1n=9

Hence, the value of is 9.

Page No 18.40:

Question 39:

if the seventh term from the beginning and end in the binomial expansion of 23+133n are equal, find n.

Answer:

In the binomail expansion of 23+133nn+1-7+1th i.e., (n − 5)th term from the beginning is the 7th term from the end.

Now,
T7=nC623n-61336=nC6×2n3-2×132And,Tn-5=nCn-6236133n-6=nC6×22×13n3-2

It is given that,
T7=Tn-5nC6×2n3-2×132=nC6×22×13n3-22n3-222=323n3-26n3-2=62n3-2=2n=12

Hence, the value of is 12.



Page No 18.45:

Question 1:

If in the expansion of (1 + x)20, the coefficients of rth and (r + 4)th terms are equal, then r is equal to
(a) 7
(b) 8
(c) 9
(d) 10

Answer:

(c) 9
Coefficients of the rth and (r+4)th terms in the given expansion are Cr-120  and 20Cr+3.Here,Cr-120  = 20Cr+3 r-1+r+3=20       if nCx=nCy   x=y or x+y=nr=2 or 2r=18r=9  

Page No 18.45:

Question 2:

The term without x in the expansion of 2x-12x212 is
(a) 495
(b) −495
(c) −7920
(d) 7920

Answer:

(d) 7920

Suppose the (r+1)th term in the given expansion is independent of x.Then, we have:Tr+1=Cr12 (2x)12-r -12x2r=(-1)r Cr12  212-2r  x12-r-2rFor this term to be independent of x, we must have:12-3r=0r=4Required term:  (-1)4 C412  212-8=12×11×10×94×3×2×16=7920

Page No 18.45:

Question 3:

If rth term in the expansion of 2x2-1x12 is without x, then r is equal to
(a) 8
(b) 7
(c) 9
(d) 10

Answer:

(c) 9
  rth term in the given expansion is Cr-112 (2x2)12-r+1 -1xr-1=(-1)r-1 Cr-112  213-r x26-2r-r+1For this term to be independent of x, we must have:27-3r=0r=9Hence, the 9th term in the expansion is independent of x.

Page No 18.45:

Question 4:

If in the expansion of (a + b)n and (a + b)n + 3, the ratio of the coefficients of second and third terms, and third and fourth terms respectively are equal, then n is
(a) 3
(b) 4
(c) 5
(d) 6

Answer:

(c) n = 5

  Coefficients of the 2nd and 3rd terms in (a+b)n are C1n and C2nCoefficients of the 3rd and 4th terms in (a+b)n+3 are C2n+3 and C3n+3Thus, we haveC1nC2n=C2n+3C3n+32n-1=3n+12n+2=3n-3n=5

Page No 18.45:

Question 5:

If A and B are the sums of odd and even terms respectively in the expansion of (x + a)n, then (x + a)2n − (xa)2n is equal to
(a) 4 (A + B)
(b) 4 (AB)
(c) AB
(d) 4 AB

Answer:

(d) 4AB
If A and B denote respectively the sums of odd terms and even terms in the expansion (x+a)nThen , (x+a)n= A+B    ...1             (x-a)n= A-B   ...2Squaring and subtraction equation 2 from1 we get (x+a)2n -(x-a)2n= A+B2- A-B2(x+a)2n -(x-a)2n=4AB

Page No 18.45:

Question 6:

The number of irrational terms in the expansion of 41/5+71/1045 is
(a) 40
(b) 5
(c) 41
(d) none of these

Answer:

(c) 41

The general term Tr+1 in the given expansion is given byCr45 (41/5)45-r (71/10)rFor Tr+1 to be an integer, we must have r5 and r10as integers i.e. 0r45 r =0,10, 20, 30 and 40Hence, there are 5 rational and 41, i.e., 46-5, irrational terms.

Page No 18.45:

Question 7:

The coefficient of x-17 in the expansion of x4-1x315 is
(a) 1365
(b) −1365
(c) 3003
(d) −3003

Answer:

(b) −1365

Suppose the (r+1)th term in the given expansion contains the coefficient of x-17.Then, we have:Tr+1=Cr15 (x4)15-r -1x3r         =(-1)r Cr15  x60-4r-3rFor this term to contain x-17 , we must have:60-7r=-17 7r=77r=11Required coefficient=(-1)11 C1115=-15×14×13×124×3×2=-1365

Page No 18.45:

Question 8:

In the expansion of x2-13x9, the term without x is equal to
(a) 2881

(b) -28243

(c) 28243

(d) none of these

Answer:

(c) 28243

Suppose the (r + 1)th term in the given expansion is independent of x.
Then , we have:
Tr+1=Cr9 (x2)9-r -13xr         =(-1)r Cr9  13rx18-2r-rFor this term to be independent of x, we must have:18-3r=0r=6Required term=(-1)6  C69  136=9×8×73×2×136=28243

Page No 18.45:

Question 9:

If an the expansion of 1+x15, the coefficients of 2r+3th and r-1th terms are equal, then the value of r is
(a) 5
(b) 6
(c) 4
(d) 3

Answer:

(a) 5
Coefficients of (2r+3)th and (r-1)th terms in the given expansion areC2r+2 15 and Cr-2.15Thus, we haveC2r+215  = Cr-2152r+2=r-2      or 2r+2+r-2=15          if nCx=nCy   x=y or x+y=n r=-4      or r=5Neglecting the negative value, We haver=5



Page No 18.46:

Question 10:

The middle term in the expansion of 2x23+32x210 is
(a) 251
(b) 252
(c) 250
(d) none of these

Answer:

(b) 252

Here, n, i.e., 10, is an even number. Middle term=102+1th term=6th termThus, we have:T6=T5+1    =C510 2x2310-5 32x25    =10×9×8×7×65×4×3×2×2535×3525    =252

Page No 18.46:

Question 11:

If in the expansion of x4-1x315, x-17 occurs in rth term, then
(a) r = 10
(b) r = 11
(c) r = 12
(d) r = 13

Answer:

(c) r = 12

Here,
Tr=Cr-115 (x4)15-r+1 -1x3r-1=(-1)r× Cr-115  x64-4r-3r+3For this term to contain x-17, we must have:67-7r=-17r=12

Page No 18.46:

Question 12:

In the expansion of x-13x29, the term independent of x is
(a) T3
(b) T4
(c) T5
(d) none of these

Answer:

(b) T4

Suppose Tr+1 is the term in the given expression that is independent of x.Thus, we have:Tr+1=Cr9 x9-r -13x2r=(-1)r Cr9  13r x9-r-2rFor this term to be independent of x, we must have9-3r=0r=3Hence, the required term is the 4th term i.e. T4

Page No 18.46:

Question 13:

If in the expansion of (1 + y)n, the coefficients of 5th, 6th and 7th terms are in A.P., then n is equal to
(a) 7, 11
(b) 7, 14
(c) 8, 16
(d) none of these

Answer:

(b) 7, 14

Coefficients of the 5th, 6th and 7th terms in the given expansion are C4n , C5n and C6nThese coefficients are in AP. Thus, we have2 C5n =C4n+C6nOn dividing both sides by C5n, we get:2=C4nC5n+C6nC5n2=5n-4+n-5612n-48=30+n2-4n-5n+20n2-21n+98=0(n-14)(n-7)=0n=7 and 14

Page No 18.46:

Question 14:

In the expansion of 12x1/3+x-1/58, the term independent of x is
(a) T5
(b) T6
(c) T7
(d) T8

Answer:

(b) T6
Suppose the (r + 1)th term in the given expansion is independent of x.
Thus, we have:
Tr+1=Cr8 12x1/38-r (x-1/5)r=Cr8 128-r x8-r3-r5For this term to be independent of x, we must have8-r3-r5=040-5r-3r=0r=5Hence, the required term is the 6th term, i.e. T6

Page No 18.46:

Question 15:

If the sum of odd numbered terms and the sum of even numbered terms in the expansion of x+an are A and B respectively, then the value of x2-a2n is
(a) A2-B2
(b) A2+B2
(c) 4 AB
(d) none of these

Answer:

(a) A2-B2
If A and B denote respectively the sums of odd terms and even terms in the expansion (x+a)nThen , (x+a)n= A+B    ...1             (x-a)n= A-B   ...2Multplying both the equations we get (x+a)n (x-a)n=A2-B2(x2-a2)n=A2-B2  

Page No 18.46:

Question 16:

If the coefficient of x in x2+λx5 is 270, then λ=
(a) 3
(b) 4
(c) 5
(d) none of these

Answer:

(a) 3

The coefficient of x in the given expansion  where x occurs at the (r+1)th term.We haveCr5 (x2)5-r λxr=Cr5 λr x10-2r-rFor it to contain x, we must have:10-3r=1r=3                 Coefficient of x in the given expansion:C35 λ3 =10λ3Now, we have10λ3=270λ3=27λ=3

Page No 18.46:

Question 17:

The coefficient of x4 in x2-3x210 is
(a) 405256

(b) 504259

(c) 450263

(d) none of these

Answer:

(a) 405256

Suppose x4 occurs at the (r+1)th term in the given expansion.Then, we haveTr+1=Cr10(x2)10-r -32x2r    =(-1)r Cr10  3r210-rx10-r-2rFor this term to contain x4, we must have:10-3r=4r=2Required coefficient = C210 3228=10×9×92×28=405256

Page No 18.46:

Question 18:

The total number of terms in the expansion of x+a100+x-a100 after simplification is
(a) 202
(b) 51
(c) 50
(d) none of these

Answer:

(b) 51
Here, n, i.e., 100, is even.
∴ Total number of terms in the expansion =n2+1=1002+1=51 

Page No 18.46:

Question 19:

If T2/T3 in the expansion of a+bn and T3/T4 in the expansion of a+bn+3 are equal, then n =
(a) 3
(b) 4
(c) 5
(d) 6

Answer:

(c) 5
In the expansion (a+b)n, we haveT2T3=C1n an-1×b1C2n an-2×b2In the expansion (a+b)n+3, we haveT3T4=C2n+3 an+1 b2C3n+3 an b3Thus, we haveT2T3=T3T4C1 naC2n b=C2n+3 aC3n+3 b2n-1=3n+12n+2=3n-3n=5

Page No 18.46:

Question 20:

The coefficient of 1x in the expansion of 1+xn 1+1xn is
(a) n !n-1 ! n+1 !

(b) 2n !n-1 ! n+1 !

(c) 2n !2n-1 ! 2n+1 !

(d) none of these

Answer:

(b) 2n !n-1 ! n+1 !

Coefficient of 1xin the given expansion=Coefficient of 1 in (1+x)n×Coefficient of1xin 1+1xn+Coefficient of x in (1+x)n×Coefficient of1x2 in 1+1xn=C0n×C1n +C1n × C2n=n+n×n!2n-2!=n+nnn-12=



Page No 18.47:

Question 21:

If the sum of the binomial coefficients of the expansion 2x+1xn is equal to 256, then the term independent of x is
(a) 1120
(b) 1020
(c) 512
(d) none of these

Answer:

(a) 1120

Suppose (r+1)th tem in the given expansion is independent of x.Then, we haveTr+1=Crn(2x)n-r 1xr    =Crn 2n-r xn-2rFor this term to be independent of x, we must haven-2r=0r=n/2Required term = Cn/2n 2n-n/2=n!n/2!2 2n/2We know:Sum of the given expansion = 256Thus, we have2n. 1n=256n=8Required term = 8!4!  4!24=1120

Page No 18.47:

Question 22:

If the fifth term of the expansion  a2/3+a-1n does not contain 'a'. Then n is equal to
(a) 2
(b) 5
(c) 10
(d) none of these

Answer:

(c) 10

T5=T4+1=C4n (a2/3)n-4 (a-1)4=C4n a2n-83-4For this term to be independent of a, we must have2n-83-4=02n-20=0n=10

Page No 18.47:

Question 23:

The coefficient of x-3 in the expansion of x-mx11 is
(a) -924 m7
(b) -792 m5
(c) -792 m6
(d) -330 m7

Answer:

(d) -330 m7
Let x-3 occur at (r+1)th term in the given expansion.Then, we haveTr+1=Cr11 x11-r -mxr=(-1)r×Cr11  mr x11-r-rFor this term to contain x-3, we must have11-2r=-3r=7Required coefficient =(-1)7 C711  m7                                         =-11×10×9×84×3×2 m7                                         =-330 m7

Page No 18.47:

Question 24:

The coefficient of the term independent of x in the expansion of ax+bx14 is
(a) 14! a7 b7

(b) 14!7!a7 b7

(c) 14!7!2a7 b7

(d) 14!7!3a7b7

Answer:

(c) 14!7!2a7 b7

Suppose (r+1)th term in the given expansion is independent of x.Then, we haveTr+1=Cr14 (ax)14-r bxr         =Cr14 a14-r br x14-2rFor this term to be independent of x, we must have14-2r=0r=7 Required term =C714 a14-7 b7=14!(7!)2a7 b7

Page No 18.47:

Question 25:

The coefficient of x5 in the expansion of 1+x21+1+x22+...+1+x30 
(a) 51C5
(b) 9C5
(c) 31C621C6
(d) 30C5 + 20C5

Answer:

(c) 31C621C6
We have 1+x21+1+x22+...1+x30             =1+x211+x10-11+x-1             =1x1+x31-1+x21Coefficient of x5 in the given expansion = Coefficient of x5 in 1x1+x31-1+x21                                                                 = Coefficient of x6 in 1+x31-1+x21                                                                 =31C6-21C6

Page No 18.47:

Question 26:

The coefficient of x8y10 in the expansion of (x + y)18 is
(a) 18C8
(b) 18p10
(c) 218
(d) none of these

Answer:

(a) 18C8

Suppose the (r+1)th term in the given expansion contains x8y10.Then, we haveTr+1=Cr18 x18-r yrFor the coefficient of x8y10 We haver= 10Hence, the required coefficient is C1018 or C818

Page No 18.47:

Question 27:

If the coefficients of the (n + 1)th term and the (n + 3)th term in the expansion of (1 + x)20 are equal, then the value of n is
(a) 10
(b) 8
(c) 9
(d) none of these

Answer:

(c) 9
Coefficient of (n+1)th term = Coefficient of (n+3)thWe have:Cn20=Cn+220      2n+2=20      if nCx=nCy   x=y or x+y=nn=9

Page No 18.47:

Question 28:

If the coefficients of 2nd, 3rd and 4th terms in the expansion of 1+xn, n  N are in A.P., then n =
(a) 7
(b) 14
(c) 2
(d) none of these

Answer:

(a) 7
Coefficients of the 2nd, 3rd  and 4th terms in the given expansion are:
C1n, C2n and C3nWe have:2×C2n=C1n+C3nDividing both sides by C2n, we get:2=C1nC2n+C3nC2n2=2n-1+n-236n-6=6+n2+2-3nn2-9n+14=0 n=7  n2 as 2>3 in the 4th term

Page No 18.47:

Question 29:

The middle term in the expansion of 2x3-32x22n is
(a) Cn2n

(b) -1n Cn2n x-n

(c) Cn2nx-n

(d) none of these

Answer:

(b) -1n Cn2n x-n

Here, n is evenMiddle term in the given expansion = 2n2+1th=(n+1)th term =Cn2n 2x32n-n-32x2n=(-1)n Cn2n x-n

Page No 18.47:

Question 30:

If rth term is the middle term in the expansion of x2-12x20, then r+3th term is
(a) C1420x214

(b) C1220 x22-12

(c) -C720 x, 2-13

(d) none of these

Answer:

(c) -C720 x× 2-13
Here n is even
So, The middle term in the given expansion is 202+1th=11th term
Therefore, (r + 3)th term is the 14th term.

T14=C1320 (x2)20-13 -12x13=-113 C1320 x14-13213=-C720 x2-13

Page No 18.47:

Question 31:

The number of terms with integral coefficients in the expansion of 171/3+351/2 x600 is
(a) 100
(b) 50
(c) 150
(d) 101

Answer:

(d) 101

The general term Tr+1 in the given expansion is given byCr600 (171/3)600-r (351/2x)r=Cr600 17200-r/3×35r/2 xrNow, Tr+1 is an integer if  r2 and r3 are integers for all 0r600Thus, we have r=0, 6, 12,...600  (Multiples of 6)Since, It is an A.PSo, 600=0+n-16    n=101Hence, there are 101 terms with integral coefficients.



Page No 18.48:

Question 32:

Constant term in the expansion of x-1x10 is
(a) 152
(b) −152
(c) −252
(d) 252

Answer:

(c) −252

Suppose (r + 1)th term is the constant term in the given expansion.
Then, we have:
Tr+1=Cr10 (x)10-r -1xr        =Cr10 (-1)r x10-r-rFor this term to be constant, we must have:10-2r=0r=5 Required term =-C510=-252

Page No 18.48:

Question 33:

If the coefficients of x2 and x3 in the expansion of (3 + ax)9 are the same, then the value of a is
(a) -79

(b) -97

(c) 79

(d) 97

Answer:

(d) 97
Coefficients of x2Coefficients of x3

C29 ×39-2 a2 =C39 ×39-3 a3a=C29C39×3       =9! ×3! ×6!×32! × 7! ×9!       =97

Page No 18.48:

Question 34:

Given the integers r > 1, n > 2, and coefficient of (3r)th and (r + 2)nd terms in the binomial expansion of (1 + x)2n are equal, then
(a) n = 2r
(b) n = 3r
(c) n = 2r + 1
(d) none of these

Answer:

Given r > 1 and n > 2
and coefficient of T3r = coefficient of Tr+2 is expansion of (1 + x)2n
i.e. 2nC3r-1=2nCr+1i.e. 3r-1=r+1           (using property of combination)and 2n-3r+1=r+1i.e. 2r=2 and  2n=4ri.e. r=1 and n=2r

Hence, the correct answer is option A.

Page No 18.48:

Question 35:

The two successive terms in the expansion of (1 + x)24 whose coefficients are in the ratio 1 : 4 are
(a) 3rd and 4th
(b) 4th and 5th
(c) 5th and 6th
(d) 6th and 7th

Answer:

For (1 + x)24 two successive terms have coefficients in ration 1 : 4
Let the two successive terms be (r + 1)th and (r + 2) terms
i.e. Tr+1=24Cr xr and  Tr+2=24Cr+1 xr+1 Given that   24Cr24Cr+1=14i.e. 24!r! (24-r)! × (r+1)! (24-r-1)!(24)!=14i.e. (r+1)r! (23-r)!r! (24-r) (23-r)!=14i.e. r+124-r=14i.e. 4r+4=24-ri.e. 5r=20i.e. r=4i.e. Tr+1=T4+1=T5 and Tr+2=T4+2=T6  are the two terms
Hence 5th  and 6th terms.
Hence, the correct answer is option C.

Page No 18.48:

Question 36:

If the coefficients of 2nd, 3rd and the 4th terms in the expansion of (1 + x)n are in A.P., then the value of n is
(a) 2
(b) 7
(c) 11
(d) 14

Answer:

Given coefficient of 2nd, 3rd and 4th terms in the expansion (1 + x)n are A.P
Since coefficient of (r + 1)th terms is nCr
∴ 2nd, 3rd and 4th coefficient are such that 2nCr = nC1 + nC3
i.e.       2×n(n-1)2= n+n(n-1)(n-2)6i.e.      n(n-1) =n1+n-1(n-2)6i.e.      n-1= 1+n-1(n-2)6i.e.      6(n-1)=6+n2-3n+2i.e.       n2-9n+14=0i.e.       n2-7n-2n+14=0i.e.      n(n-7)-2(n-7)=0i.e.      n=7          n=2 is not possible since for n=2, 3rd and 4th terms will be 0

Hence, the correct answer is option B.

Page No 18.48:

Question 37:

If the middle term of 1x+x sin x10 is equal to 778, then the value of x is
(a) 2nπ+π6

(b) nπ+π6

(c) nπ+-1nπ6

(d) nπ+-1nπ3

Answer:

Given middle term of 1x+x sinx10 is 778
Since n = 10
i.e. middle term is n2+1th term is 6th term.
 T6= T5+1=10C51x10-5(x sinx)5i.e. 638=10C51x5×x5(sinx)5i.e. 638=10!5! 5! sin5xi.e. 638=252 sin5xi.e. sin5x=132= 125i.e. sinx=12i.e. x=nπ+(-1)n π6

Hence, the correct answer is option C.

Page No 18.48:

Question 38:

The total number of terms in the expansion of (x + a)51 – (xa)51 after simplification is
(a) 102
(b) 25
(c) 26
(d) none of these

Answer:

for  (x + a)51  (x  a)51 Since (x+a)51= 51C0  x51+51C1  x50a+51C2  x49a2+ ....+51C51 a51and (x-a)51= 51C0  x51-51C1  x50a+51C2  x49a2 .... 51C51  a51Subtracting above values,(x+a)51-(x-a)51= 2 C151  x50a+C351  xa348+ ...+C5151 a51 
i.e.      1st, 3rd, 5th, 7th ________  49th, 51th term are there
  applying A.P.           
a + ( n – 1)d = 51
i.e(n – 1) 51
i.e2(n – 150
i.e26

 
Hencethe correct answer is option C.

Page No 18.48:

Question 39:

If the coefficients of x7 and x8 in 2+x3n are equal, then n is
(a) 56
(b) 55
(c) 45
(d) 15

Answer:

In (a+x)n  Tr+1=nCran-rxr In 2+x3n
Coefficient of x7 is
T8= nC7(2)n-7x37T8= nC72n-7 x737
and coefficient of x8 is
T9= nC8(2)n-8x38T9= nC82n-8 x838
Since T8 = T9 i.e coefficient of T8 = coefficient of T9
 nC72n-737=nC8 2n-838n!   2n-77!(n-7)!×37=n! × 2n-88!(n-8)!×38i.e.  8! 7!  (n-8)! (n-7)(n-8)! = 2n-83×2n-7i.e.  8n-7= 16 48= n-7i.e.  n= 55

Hence, the correct answer is option B.

Page No 18.48:

Question 40:

The ratio of the coefficient of x15 to the term independent of x in x2+2x15, is
(a) 12 : 32
(b) 1 : 32
(c) 32 : 12
(d) 32 : 1

Answer:

In x2+2x15 we have Tr+1=15Cr(x2)15-r2xr
i.e. General term is  Tr+1=15Cr x30-3r 2r
Hence for the term independent of x,
30 – 3r = 0
i.e. r = 10
hence T11 has coefficient 15C10 210       ...(1)
and term with x15 will have 30 – 3r = 15
i.e. 15 = 3r
i.e. r = 5
∴ coefficient will be 15C5 25              ...(2)
∴  ratio of coefficient of x15 to the term independent of x will be
 eqn(2)eqn(1)=25×15C5 210×15C10 =125           15C10=15C5-10=15C5
i.e. ratio will be 1 : 32
Hence, the correct answer is option B.

Page No 18.48:

Question 41:

If z=32+i25+32-i25, then
(a) Re (z) = 0
(b) Im (z) = 0
(c) Re (z) > 0, Im (z) > 0
(d) Re (z) > 0, Im (z) < 0

Answer:

for  32+i2r=34+14=44=1  and θ=tan-11232=tan-113i.e. θ=π6i.e. 32+i2=r(cosθ+isinθ)we have  θ=π6 and r=1and  32-i2=rcosθ-isinθ

 z=32+i25+32-i25= rcosθ+isinθ5+r(cosθ-isinθ)5 where r=1, θ=π6= cosθ+isinθ5+cosθ-isinθ5= cos5θ+isin5θ+cos5θ-isin5θ      (By De-moivre theorem)= 2 cos5θ

Page No 18.48:

Question 42:

If (1 – x + x2)n = a0 + a1x + a2x2 +...+a2n  x2n, then a0 + a2 + a4 +...+ a2n equals

(a) 3n+12

(b) 3n-12

(c) 1-3n2

(d) 3n+12

Answer:

Given:
(1  x + x2)n = a0 + a1 x + a2 x2 + ....+a2n  x2n     ...(1)In order to find  a0 + a2 + a4 +....+ a2n
i.e. all even terms are involved
∴ replace x by 1 in equation (1)
we get
(1-1+1)n=a0+a1+a2+ .... +a2ni.e. 1=a0+a1+a2+ .... +a2n      ....(2)
and now replace x by –1 in equation (1), we get
(1+1+1)n=a0-a1+a2-a3+ .... +a2ni.e. 3n=a0-a1+a2-a3+ .... +a2n        ...(3)

By adding (2) and (3), we get

3n+1=2a0+2a2+2a4+....+2a2ni.e. a0+a2+a4+....+a2n=1+3n2

Hence, the correct answer is option A.



Page No 18.49:

Question 1:

The largest coefficient in (1 + x)30 is ___________.

Answer:

for (1+x)30Tr+1=nCr(1)30-r xrSince n=30        i.e.   evenThe largest coefficient is for r=n2=15
i.e for middle term
which is 30C15.

Page No 18.49:

Question 2:

The largest coefficient in (1 + x)41 is ___________.

Answer:

In (1 + x)41
Since n is odd i.e 41
∴ The largest coefficient is 41C21 or 41C20
i.e. n+12 or n+12+1th term gives largest coefficient

Page No 18.49:

Question 3:

The number of terms in the expansion of (x + y + z)n is ___________.

Answer:

(x+y+z)n  =[x+(y+z)]n = nC0 xn+nC1 xn-1(y+z)+nC2 xn-2(y+z)2+....+nCn(y+z)n
∴ Number of terms in the expansion
= 1+2+....+n+(n+1)=(n+1)(n+2)2= n+2C2

Page No 18.49:

Question 4:

Middle term in the expansion of (a3 + ba)28 is ___________.

Answer:

In (a3 + ba)28
Here n = 28
So, there is one middle term
i.e. 282+1th term which is 15th term Middle term is T15

i.e.  T14+1=28C14a328-14 (ba)14     using Tr+1=nCr xryn-r for (x+y)n= 28C14a314 b14 a14= 28C14 a56 a14

Page No 18.49:

Question 5:

The ratio of the coefficients of xm and xn in the expansion of (1 + x)m + n is ___________.

Answer:

For (1 + x)m + n
Tr+1 = m+nCrxr
for coefficient of xm, r = m
i.e coefficient is m+nCm
and for coefficient of xn, r = n
i.e. coefficient is m+nCn
 Ratio of coefficient of xm to xn is m+nCmm+nCn=m+nCm+n-mm+nCn=m+nCnm+nCn=1

Page No 18.49:

Question 6:

The coefficient of a–6b4 in the expansion 1a-23b10 is ___________.

Answer:

In  1a-23b10Tr+1=10Cr1a10-r-2b3rfor coefficient of  a-6 b4put 10-r=6i.e. r=4
We get
coefficient of a-6 b4= 10C4 -234=10!4! 6!×2434=10×9×8×72×3×4×2434=10×7×2433

i.e. coefficient of a-6 b4 =112027.

Page No 18.49:

Question 7:

In the expansion of x2-1x216, the value of the constant term is ___________.

Answer:

In x2 - 1x216Tr+1=Cr16 x216-r-1x2r        = Cr16 x32-2r -1r x-2ri.e. Tr+1= Cr16 -1r x32-4rfor constant term, x32-4r = x0i.e. 32 - 4r=0i.e. r=8  T8+1=C816
i.e value of constant term is 16C8.

Page No 18.49:

Question 8:

The position of the term independent of x in the expansion of x3+32x210 is ___________.

Answer:

for x3+32x210Tr+1 =10Cr x310-r 32x2r         = 10Cr1310-r x10-r2 x-2r 32rTr+1 =10Cr310-r 32r  x10-5r2 for constant term, 10-5r2=0     i.e. r=2
Hence, third term is independent of x.

Page No 18.49:

Question 9:

If 215 is divided by 13, the remainder is ___________.

Answer:

If 215 is divided by 13
Since 215 = (5 – 3)15
= 10C0 (5)15 (–3)0 + ..........
Correction

Page No 18.49:

Question 10:

The sum of the series r=01020Cr is ___________.

Answer:

for  r=010  20Crfor n=2consider,(1+x)20= 20C0+ 20C1x+......+ 20C20x20i.e.  r=010  20Cr= 20C0+ 20C1+ 20C2+......+ 20C10Since  20C1= 20C19             20C2= 20C18                  .                  .                  .             20C11= 20C9i.e. (1+x)20=2( 20C0+ 20C1x+.......+ 20C9x9+ 20C10x10)- 20C10x10       (1+x)20=2  r=010  20Crxr-20C10x10                         2  r=010  20Crxrput x=1i.e. 2 r=010  20Cr=(1+1)20+20C10i.e.  r=010  20Cr=219+20C102

Page No 18.49:

Question 11:

The number of terms in the expansion of 2x+y347 is ___________.

Answer:

In {(2x + y3)4}7
n = ??
Since In (a + b)n ; number of terms is n + 1
∴ {(2x + y3)}28
Number of terms is 28 + 1 = 29

Page No 18.49:

Question 12:

The middle term in the expansion of x-1x18 is ___________.

Answer:

x-1x18
Since 18 is even
there is only one middle term i.e. 2n2+1thterm
i.e. (n + 1)th term
i.e. Tn+1= 2nCn(x)2n-n -1xn= 2nCn (-1)n x2n-n x-n= 2nCn (-1)n x0     i.e. 18C9 (-1)18=  18C9

Page No 18.49:

Question 13:

The coefficient of the middle term in the expansion of (1 + x)10 is ___________.

Answer:

In (1 + x)10
Tr +1 = 10Cr xr
Middle term is obtained when r = 5
i.e. Tr +1 = 10C5x5
i.e. coefficient of middle term is 10C5

Page No 18.49:

Question 14:

The total number of terms in the expansion of (1 + x)2n – (1 – x)2n ___________.

Answer:

In (1+x)2n-(1-x)2nSince(1+x)2n= 2nC0+ 2nC1x+.....+ 2nC2nx2n(1-x)2n= 2nC0- 2nC1x+......+ 2nC2nx2n 
Subtracting above two,
i.e. (1+x)2n-(1-x)2n=2 2nC1x+ 2nC3x3+.... 2nC2n-1x2n-1
i.e. number of terms here=2×n2 (odd from 2n is n)=n

Page No 18.49:

Question 15:

If x4 occurs in the rth terms in the expansion of x4+1x315, then r = ___________.

Answer:

for x4 + 1x315
n = 15
Tr+1 = C415 x4r 1x315-r= C415 x4r x-45-3r= C4 15x4r x3r-45= C4 15 x7r-45
To get x4, 7r – 45 = 4
i.e. 7r = 49
i.e r = 7

Page No 18.49:

Question 16:

The coefficient of x in the binomial expansion of x2+ax5 is ___________.

Answer:

for space space open parentheses x squared plus a over x close parentheses to the power of 5
T subscript r plus 1 end subscript equals space to the power of 5 C subscript r space end subscript open parentheses x squared close parentheses to the power of r space open parentheses a over x close parentheses to the power of 5 minus r end exponent
space space space space space space space space space equals space to the power of 5 C subscript r space space end subscript x to the power of 2 r end exponent space a to the power of 5 minus r end exponent space x to the power of r minus 5 end exponent
T subscript r plus 1 end subscript equals space to the power of 5 C subscript r space end subscript a to the power of 5 minus r end exponent space x to the power of 2 r plus r minus 5 end exponent
For coefficient  of x, put 3r – 5 = 1
i.e. 3r = 6
i.e. r = 2
i.e. coefficient x is 5C2a3 = 10 a3



Page No 18.50:

Question 17:

If A and B are the coefficient of xn in the expansion of (1 + x)2n and (1 + x)2n – 1 respectively, then AB=___________.

Answer:

The coefficient of xn is (1 + x)2n is ?
Since Tr +1 = 2nCr xr
For xn coefficient put r = n
i.e coefficient of xn is 2nC
i.e. A = 2nCn
and for coefficient of xn in (1 + x)2n–1
Tr +1 = 2n–1Cr xr
Put r = n
i.e. coefficient of xn in (1 + x)2n–1 is 2n–1Cn
i.e B = 2n–1Cn
 AB= 2nCn 2n-1Cn=(2n)! n! n!n! n! (2n-1)!=2

Page No 18.50:

Question 18:

If the coefficients of x7 and x8 in 2+x3nare equal, then n = ___________.

Answer:

In space space space open parentheses 2 plus x over 3 close parentheses to the power of n
T subscript r plus 1 end subscript equals space to the power of n C subscript r space end subscript open parentheses x over 3 close parentheses to the power of r space left parenthesis 2 right parenthesis to the power of n minus r end exponent
T subscript r plus 1 end subscript equals space fraction numerator blank to the power of n C subscript r space 2 to the power of n minus r end exponent over denominator 3 to the power of r end fraction space x to the power of r
therefore space space coefficient space of space x to the power of 7 space is space fraction numerator blank to the power of n C subscript 7 space 2 to the power of n minus 7 end exponent over denominator 3 to the power of 7 end fraction
and space coefficient space of space x to the power of 8 space is space fraction numerator blank to the power of n C subscript 8 space 2 to the power of n minus 8 end exponent over denominator 3 to the power of 8 end fraction
According to given condition,
Coefficient of x7 = Coefficient of x8
straight i. straight e. space fraction numerator blank to the power of n C subscript 7 space 2 to the power of n minus 7 end exponent over denominator 3 to the power of 7 end fraction equals fraction numerator blank to the power of n C subscript 8 space 2 to the power of n minus 8 end exponent over denominator 3 to the power of 8 end fraction
space space space space space space fraction numerator n factorial over denominator 7 factorial left parenthesis n minus 7 right parenthesis factorial end fraction cross times 2 to the power of n minus 7 end exponent over 3 to the power of 7 space equals space space fraction numerator n factorial over denominator 8 factorial left parenthesis n minus 8 right parenthesis factorial end fraction cross times 2 to the power of n minus 8 end exponent over 3 to the power of 8
straight i. straight e. space fraction numerator 8 cross times 7 factorial over denominator 7 factorial space end fraction cross times fraction numerator left parenthesis n minus 8 right parenthesis factorial over denominator left parenthesis n minus 7 right parenthesis left parenthesis n minus 8 right parenthesis factorial end fraction space equals space 2 to the power of n minus 8 end exponent over 2 to the power of n minus 7 end exponent cross times 3 to the power of 7 over 3 to the power of 8
space space space space space space space fraction numerator 8 over denominator n minus 7 end fraction space equals space space fraction numerator 1 over denominator 2 cross times 3 end fraction equals 1 over 6 space
straight i. straight e. space 48 equals space n minus 7
straight i. straight e. space n equals space 55

Page No 18.50:

Question 19:

If 13th term in the expansion of x2+2xn is independent of x, then the value of n is ___________.

Answer:

In x2 + 2xnTr+1 = Crn x2n-r 2xr          = Crn x2n-2r 2rxr          = Crn x2n-2r x-r  2rTr+1= Cr n2r x2n-3r
If for r = 12, i.e 13th term is independent of x
2n – 3r = 0
⇒ 2n = 3 × 12
i.e. n = 18

Page No 18.50:

Question 20:

The term independent of x in the expansion of x+1x10 is ___________.

Answer:

In space open parentheses square root of x plus fraction numerator 1 over denominator square root of x end fraction close parentheses to the power of 10
T subscript r plus 1 end subscript space equals to the power of 10 C subscript r space left parenthesis square root of x right parenthesis to the power of 10 minus r end exponent space open parentheses fraction numerator 1 over denominator square root of x end fraction close parentheses to the power of r
space space space space space space space space space equals space to the power of 10 C subscript r space x to the power of fraction numerator 10 minus r over denominator 2 end fraction end exponent cross times space x to the power of fraction numerator negative r over denominator 2 end fraction end exponent
T subscript r plus 1 end subscript space equals to the power of 10 C subscript r space x to the power of fraction numerator 5 minus r over denominator 2 end fraction space end exponent x to the power of fraction numerator negative r over denominator 2 end fraction end exponent equals to the power of 10 C subscript r space space x 5 minus r space
For tern to be independent of x, 5 – r = 0
i.e r = 5
Hence space term space is space to the power of 10 C subscript 5 space equals fraction numerator 10 cross times 9 cross times 8 cross times 7 cross times 6 over denominator 5 cross times 4 cross times 3 cross times 2 end fraction equals 252

Page No 18.50:

Question 1:

Write the number of terms in the expansion of 2+3x10+2-3x10.

Answer:

Number of terms in the expansion (x+y)n+(x-y)n where n is even=n2+1Thus, we have:Number of terms in the given expansion=102+1=6

Page No 18.50:

Question 2:

Write the sum of the coefficients in the expansion of 1-3x+x2111.

Answer:

To find the sum of coefficients, we plug 1 for each variable then, we get the sum of coefficients of the given expression. Sum of coefficient =1-3x+x2111                              =1-3×1+12111                              =1-3+1111                              =1-3+1111                              =-1111                              =-1

Page No 18.50:

Question 3:

Write the number of terms in the expansion of 1-3x+3x2-x38.

Answer:

The given expression is (1-3x+3x2-x3)8. It can be written as [(1-x)3]8 i.e. (1-x)24Hence, the number of terms is 24+1 i.e. 25

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Question 4:

Write the middle term in the expansion of 2x23+32x2.

Answer:

Here, n, i.e., 10, is an even number. Middle term=102+1th term=6th termThus, we have:T6=T5+1    =C510 2x2310-5 32x25    =10×9×8×7×65×4×3×2×2535×3525    =252

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Question 5:

Which term is independent of x, in the expansion of x-13x29?

Answer:

Suppose Tr+1 is the term in the given expression that is independent of x.Thus, we have:Tr+1=Cr9 x9-r -13x2r=(-1)r Cr9  13r x9-r-2rFor this term to be independent of x, we must have9-3r=0r=3Hence, the required term is the 4th term.

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Question 6:

If a and b denote respectively the coefficients of xm and xn in the expansion of 1+xm+n, then write the relation between a and b.

Answer:

  Coefficient of xm in the given expansion = Cmm+n =aCoefficient of xn in the given expansion = Cnm+n =ba=b   Cmm+n =Cnm+n

Page No 18.50:

Question 7:

If a and b are coefficients of xn in the expansions of 1+x2n and 1+x2n-1 respectively, then write the relation between a and b.

Answer:

Coefficient of xn in the expansion (1+x)2n = Cn2n=aCoefficient of xn in the expansion (1+x)2n-1=Cn2n-1=bNow, we have: Cn2n=2n!n!. n!=2n(2n-1)!nn-1! n!    ...1 and Cn2n-1=(2n-1)!n!(n-1)!       ...2Dividing equation 1 by 2, we get Cn2nCn2n-1=2n(2n-1)! n! (n-1)!nn-1! n! (2n-1)!ab=2a=2b



Page No 18.51:

Question 8:

Write the middle term in the expansion of x+1x10.

Answer:

Here, n, i.e., 10, is an even number. Middle term=102+1th term=6th termThus, we have:T6=T5+1     =C510 x10-5×1x5     =C510

Page No 18.51:

Question 9:

If a and b denote the sum of the coefficients in the expansions of 1-3x+10x2n and 1+x2n respectively, then write the relation between a and b.

Answer:

Here,a=1-3+10=8=23b=1+1=2a=b3

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Question 10:

Write the coefficient of the middle term in the expansion of 1+x2n.

Answer:

Here, n, i.e., 2n, is an even number. Middle term=2n2+1th term=n+1th termThus, we have:Tn+1=2nCn12n-nxn          =2nCnxnHence, the coefficient of the middle term is 2nCn

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Question 11:

Write the number of terms in the expansion of 2x+y347.

Answer:

In the binomial expansion of a+bn, total number of terms will be (n + 1).

Now, 2x+y347=2x+y328

Therefore, in the expansion of 2x+y347, total number of terms will be 28 + 1 = 29.

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Question 12:

Find the sum of the coefficients of two middle terms in the binomial expansion of 1+x2n-1.

Answer:

1+x2n-1Here, n is an odd number.Therefore, the middle terms are 2n-1+12th and 2n-1+12+1th, i.e., nth and (n+1)th terms.Now, we haveTn=Tn-1+1=Cn-12n-1 xn-1And,Tn+1=Tn+1=Cn2n-1 xn the coefficients of two middle terms are Cn-12n-1 and Cn2n-1.Now,2n-1Cn-1+2n-1Cn=2nCn

Hence, the sum of the coefficients of two middle terms in the binomial expansion of 1+x2n-1 is 2nCn.

Page No 18.51:

Question 13:

Find the ratio of the coefficients of xp and xq in the expansion of 1+xp+q.

Answer:

Coefficient of xp in the expansion of 1+xp+q is p+qCp.

Coefficient of xq in the expansion of 1+xp+q is p+qCq.

Now,
p+qCpp+qCq=p+q!p!q!p+q!q!p!=1

Hence, the ratio of the coefficients of xp and xq in the expansion of 1+xp+q is 1 : 1.

Page No 18.51:

Question 14:

Write last two digits of the number 3400.

Answer:

3400=9200     =10-1200     =200C010200+200C110199-11+.....+200C198102-1198+200C199101-1199+200C200-1200     =10010198+200C110197-11+.....+200C198-1198+200101-1199+-1200     =10010198-200C110197+.....+200C198-210+1     =100(a natural number)+1

Hence, last two digits of the number 3400 is 01.

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Question 15:

Find the number of terms in the expansion of a+b+cn.

Answer:

We have,
a+b+cn=a+b+cn                 =an+nC1 an-1b+c1+nC2 an-2b+c2+...+nCnb+cn

Further, expanding each term of R.H.S., we note that
First term consists of 1 term.
Second term on simplification gives 2 terms.
Third term on expansion gives 3 terms.
Similarly, fourth term on expansion gives 4 terms and so on.

∴ The total number of terms = 1 + 2 + 3 + .... + (n + 1) = n+1n+22.

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Question 16:

If a and b are the coefficients of xn in the expansion of 1+x2n and 1+x2n-1 respectively, find ab.

Answer:

Coefficients of xn in the expansion of 1+x2n is 2nCn=a.

Coefficients of xn in the expansion of 1+x2n-1 is 2n-1Cn=b.

Now,
ab=2nCn2n-1Cn     =2n!n!n!2n-1!n!n-1!     =2nn     =2

Hence, ab=2.

Page No 18.51:

Question 17:

Write the total number of terms in the expansion of x+a100+x-a100.

Answer:

The total number of terms are 101 of which 50 terms get cancelled.

Hence, the total number of terms in the expansion of x+a100+x-a100 is 51.

Page No 18.51:

Question 18:

If 1-x+x2n=a0+a1 x+a2 x2+...+a2n x2n, find the value of a0+a2+a4+...+a2n.

Answer:

Putting x = 1 and −1 in
1-x+x2n=a0+a1 x+a2 x2+...+a2n x2n
we get,

1=a0+a1+a2+...+a2n       ...(1)
and
3n=a0-a1+a2-...+a2n       ...(2)

Adding (1) and (2), we get
3n+1=2a0+a2+...+a2n

Hence, the value of a0+a2+a4+...+a2n is 3n+12.



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