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^{2})(x-(3/x))^{11}^{2}+6n-1)/3WHAT CAN I DO?

PLZZZZZ GUIDE ME....

$\left(A\right)zero\phantom{\rule{0ex}{0ex}}\left(B\right)1\phantom{\rule{0ex}{0ex}}\left(C\right)2\phantom{\rule{0ex}{0ex}}\left(C\right)noneofthese$

^{n }is 1.NO LINKS PLEASExand^{2}xin the binomial expansion of^{11}(1+ax+2xbe^{2})and27respectively, show that,(-192)a= -1NO LINKS PLEASEa] 6

b] 9

c] 12

d] 15

.P(n) be astatement:2^{n}<n, neN .Show that the statement is not true for any n.The solution stated that at least 2 means 2 or more than 2. But the expression written for this purpose subtracts

2n(x intersect y intersect z)Pl explain why

2n(x intersect y intersect z)is being subtractedQ- find the coefficient of x^20 in (1+3x+3x^2+x^3).

Sir, you did this question in live class of B.T (I) at 51:16 , you took (x+1)^60 that i got but solve it from (1+x)^60 as my answer is coming ^60C_20 from that. Thanks

Hello sir this is Noel Saji a new student for jee, i have a doubt in this question

(x+1/x2/3-x1/3 - x-1/x-x1/2 )

Q13. If $3.{}^{n}C_{0}-8.{}^{n}C_{1}+13.{}^{n}C_{2}-18.{}^{n}C_{3}+.....$ upto $\left(n+1\right)$ terms is $\left(n\ge 2\right)$:

(A) zero

(B) 1

(C) 2

(D) none of these

1^3+2^3+3^3+__________+n^3 = [n(n+1)/2]^2