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#### Question 1:

1 + 2 + 3 + ... + n$\frac{n\left(n+1\right)}{2}$ i.e. the sum of the first n natural numbers is $\frac{n\left(n+1\right)}{2}$.

Let P(n) be the given statement.
Now,
P(n) = 1 + 2 + 3 +...+ n  =$\frac{n\left(n+1\right)}{2}$

By the principle of mathematical induction, P(n) is true for all n$\in$N.

#### Question 2:

12 + 22 + 32 + ... + n2 =

Let P(n) be the given statement.
Now,

#### Question 3:

1 + 3 + 32 + ... + 3n−1 = $\frac{{3}^{n}-1}{2}$

Let P(n) be the given statement.
Now,

#### Question 4:

$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}=\frac{n}{n+1}$

Let P(n) be the given statement.
Now,

#### Question 5:

1 + 3 + 5 + ... + (2n − 1) = n2 i.e., the sum of first n odd natural numbers is n2.

Let P(n) be the given statement.
Now,

#### Question 6:

$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}=\frac{n}{6n+4}$

Let P(n) be the given statement.
Now,

#### Question 7:

$\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{\left(3n-2\right)\left(3n+1\right)}=\frac{n}{3n+1}$

Let P(n) be the given statement.
Now,

#### Question 8:

$\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2n+1\right)\left(2n+3\right)}=\frac{n}{3\left(2n+3\right)}$

Let P(n) be the given statement.
Now,

#### Question 9:

$\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.5}+...+\frac{1}{\left(4n-1\right)\left(4n+3\right)}=\frac{n}{3\left(4n+3\right)}$

Let P(n) be the given statement.
Now,

#### Question 10:

1.2 + 2.22 + 3.23 + ... + n.2n = (n − 1) 2n+1+2

Let P(n) be the given statement.
Now,

#### Question 11:

2 + 5 + 8 + 11 + ... + (3n − 1) =$\frac{1}{2}n\left(3n+1\right)$

Let P(n) be the given statement.
Now,

#### Question 12:

1.3 + 2.4 + 3.5 + ... + n. (n + 2) =$\frac{1}{6}n\left(n+1\right)\left(2n+7\right)$

Let P(n) be the given statement.
Now,

#### Question 13:

1.3 + 3.5 + 5.7 + ... + (2n − 1) (2n + 1) = $\frac{n\left(4{n}^{2}+6n-1\right)}{3}$

Let P(n) be the given statement.
Now,

#### Question 14:

1.2 + 2.3 + 3.4 + ... + n (n + 1) = $\frac{n\left(n+1\right)\left(n+2\right)}{3}$

Let P(n) be the given statement.
Now,

#### Question 15:

$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{{2}^{n}}=1-\frac{1}{{2}^{n}}$

Let P(n) be the given statement.
Now,

#### Question 16:

12 + 32 + 52 + ... + (2n − 1)2 = $\frac{1}{3}n\left(4{n}^{2}-1\right)$

Let P(n) be the given statement.
Now,

#### Question 17:

a + ar + ar2 + ... + arn−1 = $a\left(\frac{{r}^{n}-1}{r-1}\right),r\ne 1$

Let P(n) be the given statement.
Now,

#### Question 18:

a + (a + d) + (a + 2d) + ... (a + (n − 1) d) = $\frac{n}{2}\left[2a+\left(n-1\right)d\right]$

Let P(n) be the given statement.
Now,

#### Question 19:

52n −1 is divisible by 24 for all nN.

Let P(n) be the given statement.
Now,

#### Question 20:

32n+7 is divisible by 8 for all nN.

Let P(n) be the given statement.
Now,

#### Question 21:

52n+2 −24n −25 is divisible by 576 for all nN.

Let P(n) be the given statement.
Now,

#### Question 22:

32n+2 −8n − 9 is divisible by 8 for all nN.

Let P(n) be the given statement.
Now,

#### Question 23:

(ab)n = anbn for all nN.

Let P(n) be the given statement.
Now,

#### Question 24:

n(n + 1) (n + 5) is a multiple of 3 for all nN.

Let P(n) be the given statement.
Now,

#### Question 25:

72n + 23n−3. 3n−1 is divisible by 25 for all nN.

Let P(n) be the given statement.
Now,

#### Question 26:

2.7n + 3.5n − 5 is divisible by 24 for all nN.

Let P(n) be the given statement.
Now,

#### Question 27:

11n+2 + 122n+1 is divisible by 133 for all nN.

Let P(n) be the given statement.
Now,

#### Question 28:

Given for n ≥ 2, where a > 0, A > 0.
Prove that $\frac{{a}_{n}-\sqrt{A}}{{a}_{n}+\sqrt{A}}=\left(\frac{{a}_{1}-\sqrt{A}}{{a}_{1}+\sqrt{A}}\right){2}^{n-1}$

#### Question 29:

Prove that n3 $-$ 7+ 3 is divisible by 3 for all n $\in$ N.

Hence, n3 $-$ 7+ 3 is divisible by 3 for all n $\in$ N.

#### Question 30:

Prove that 1 + 2 + 22 + ... + 2n = 2n+1 $-$ 1 for all n $\in$ N.

Hence, 1 + 2 + 22 + ... + 2n = 2n+1 $-$ 1 for all n $\in$ N.

#### Question 31:

7 + 77 + 777 + ... + 777  $\underset{n-\mathrm{digits}}{...........}7=\frac{7}{81}\left({10}^{n+1}-9n-10\right)$

Let P(n) be the given statement.
Now,

Now, P(m + 1) = 7 + 77 + 777 +....+ 777...(m + 1) digits...7

#### Question 32:

$\frac{{n}^{7}}{7}+\frac{{n}^{5}}{5}+\frac{{n}^{3}}{3}+\frac{{n}^{2}}{2}-\frac{37}{210}n$ is a positive integer for all n ∈ N.

Let P(n) be the given statement.
Now,

#### Question 33:

$\frac{{n}^{11}}{11}+\frac{{n}^{5}}{5}+\frac{{n}^{3}}{3}+\frac{62}{165}n$ is a positive integer for all nN.

Let P(n) be the given statement.
Now,

#### Question 34:

for all nN and $0

We need to prove for all nN and $0 using mathematical induction.

For n = 1,

LHS = $\frac{1}{2}\mathrm{tan}\frac{x}{2}$

and

Therefore, the given relation is true for n = 1.

Now, let the given relation be true for n = k.

We need to prove that the given relation is true for n = k + 1.

Now,

Let:

.

Now,

Thus, for all nN and $0.

#### Question 35:

Let P(n) be the statement : 2n ≥ 3n. If P(r) is true, show that P(r + 1) is true. Do you conclude that P(n) is true for all nN?

#### Question 36:

$\frac{\left(2n\right)!}{{2}^{2n}\left(n!{\right)}^{2}}\le \frac{1}{\sqrt{3n+1}}$ for all nN

Let P(n) be the given statement.
Thus, we have:

Now,

$⇒\frac{\left(2m+2\right)!}{{2}^{2m+2}{\left(\left(m+1\right)!\right)}^{2}}\le \sqrt{\frac{{\left(2m+1\right)}^{2}}{4{\left(m+1\right)}^{2}\left(3m+1\right)}}\phantom{\rule{0ex}{0ex}}⇒\frac{\left(2m+2\right)!}{{2}^{2m+2}{\left(\left(m+1\right)!\right)}^{2}}\le \sqrt{\frac{\left(4{m}^{2}+4m+1\right)×\left(3m+4\right)}{4\left(3{m}^{3}+7{m}^{2}+5m+1\right)\left(3m+4\right)}}\phantom{\rule{0ex}{0ex}}⇒\frac{\left(2m+2\right)!}{{2}^{2m+2}{\left(\left(m+1\right)!\right)}^{2}}\le \sqrt{\frac{12{m}^{3}+28{m}^{2}+19m+4}{\left(12{m}^{3}+28{m}^{2}+20m+4\right)\left(3m+4\right)}}\phantom{\rule{0ex}{0ex}}\because \frac{12{m}^{3}+28{m}^{2}+19m+4}{\left(12{m}^{3}+28{m}^{2}+20m+4\right)}<1\phantom{\rule{0ex}{0ex}}\therefore \frac{\left(2m+2\right)!}{{2}^{2m+2}{\left(\left(m+1\right)!\right)}^{2}}<\frac{1}{\sqrt{3m+4}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Thus, P(m + 1) is true.

Hence, by mathematical induction $\frac{\left(2n\right)!}{{2}^{2n}\left(n!{\right)}^{2}}\le \frac{1}{\sqrt{3n+1}}$ is true for all nN

#### Question 37:

$1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{{n}^{2}}<2-\frac{1}{n}$for all n ≥ 2, nN

Let P(n) be the given statement.
Thus, we have:

#### Question 38:

x2n−1 + y2n−1 is divisible by x + y for all nN.

Let P(n) be the given statement.
Now,

#### Question 39:

Let P(n) be the given statement.

#### Question 40:

[NCERT EXEMPLAR]

#### Question 41:

[NCERT EXEMPLAR]

#### Question 44:

Show by the Principle of Mathematical induction that the sum Sn of then terms of the series ${1}^{2}+2×{2}^{2}+{3}^{2}+2×{4}^{2}+{5}^{2}+2×{6}^{2}+{7}^{2}+...$ is given by

[NCERT EXEMPLAR]

#### Question 45:

Prove that the number of subsets of a set containing n distinct elements is 2n, for all n $\in$ N.                                         [NCERT EXEMPLAR]

Hence, the number of subsets of a set containing n distinct elements is 2n, for all n $\in$ N.

#### Question 46:

[NCERT EXEMPLAR]

#### Question 47:

[NCERT EXEMPLAR]

Disclaimer: It should be k instead n in the denominator of ${x}_{k}=\frac{{x}_{k-1}}{k}$. The same has been corrected above.

#### Question 48:

[NCERT EXEMPLAR]

#### Question 49:

[NCERT EXEMPLAR]

#### Question 1:

Make the correct alternative in following question:

If xn $-$ 1 is divisible by x $-$ $\lambda$, then the least positive integral value of $\lambda$ is

(a) 1                    (b) 2                    (c) 3                    (d) 4

Hence, the correct alternative is option (a).

#### Question 2:

Make the correct alternative in the following question:

(a) 19                         (b) 17                         (c) 23                         (d) 25

Hence, the correct alternative is option (b).

#### Question 3:

Make the correct alternative in the following question:

If ${10}^{n}+3×{4}^{n+2}+\lambda$ is divisible by 9 for all n $\in$ N, then the least positive integral value of $\lambda$ is

(a) 5                         (b) 3                         (c) 7                         (d) 1

Hence, the correct alternative is option (a).

#### Question 4:

Make the correct alternative in the following question:

Let P(n): 2n < (1 × 2 × 3 × ... × n). Then the smallest positive integer for which P(n) is true

(a) 1                         (b) 2                        (c) 3                        (d) 4

As, 2n < (1 × 2 × 3 × ... × n) is possible only when n $\ge$ 4

So, the smallest positive integer for which P(n) is true, is 4.

Hence, the correct alternative is option (d).

#### Question 5:

Make the correct alternative in the following question:

A student was asked to prove a statement P(n) by induction. He proved P(k +1) is true whenever P(k) is true for all > 5 $\in$ N and also P(5) is true. On the basis of this he could conclude that P(n) is true.

(a) for all n $\in$ N                         (b) for all > 5                         (c) for all n $\ge$ 5                         (d) for all n < 5

As, P(5) is true and
P(k + 1) is true whenever P(k) is true for all k > 5 $\in$ N.

By the definition of the priniciple of mathematical induction, we get
P(n) is true for all n $\ge$ 5.

Hence, the correct alternative is option (c).

#### Question 6:

Make the correct alternative in the following question:

If P(n): 49n + 16n + $\lambda$ is divisible by 64 for n $\in$ N is true, then the least negative integral value of $\lambda$ is

(a) $-$3                         (b) $-$2                         (c) $-$1                         (d) $-$4

Hence, the correct alternative is option (c).

#### Question 1:

If P(n) : "2 × 42n + 1 + 33n + 1 is divisible by λ for all nN" is true, then the value of λ is _____________.

If P(n) = 2 × 42n + 1 + 33n + 1 is divisible by λ ∀ n ∈ N is true

for n = 1,

for n = 2,

Since common factor of P(1) and P(2) is 11
Hence 2 × 42n+1 + 33n+1 is divisible by 11
i.e λ = 11

#### Question 2:

If P(n) : 2n < n!, nN, then P(n) is true for all n ≥ _____________.

Given P(n) : 2n < n! ; n ∈ N

for n = 1,
P(1) : 2' < 1!
i.e 2 < 1
Which is not true

for n = 2,
P(2) : 22 = 4 < 2!
i.e 4 < 2
Which is not true

for n = 3,
P(3) : 23 < 3!
i.e 8 < 1 × 2 × 3
i.e 8 < 6
Which is again not true.

for n = 4,
P(4) i.e 24 < 4!
i.e 16 < 24
i.e a true statement.

P(5) : 25 < 5!
i.e 32 < 1 × 2 × 3 × 4 × 5
i.e 32 < 120
Which is also true
P(n) : 2n < n! is true for n > 3
P(n) : 2n < n! is true i.e for n ≥ 4

#### Question 3:

If P(n) : 2n < n!, nN, then P(n) is true for all n > _____________.

Given P(n) : 2n < n! ; nN,

for n = 1,
P(1) : 2' < 1!
i.e 2 < 1
Which is not true

for n = 2,
P(2) : 22 = 4 < 2!
i.e 4 < 2
Which is not true

for n = 3,
P(3) : 23 < 3!
i.e 8 < 1 × 2 × 3
i.e 8 < 6
Which is again not true.

for n = 4,
P(4) i.e 24 < 4!
i.e 16 < 24
i.e a true statement.

P(5) : 25 < 5!
i.e 32 < 1 × 2 × 3 × 4 × 5
i.e 32 < 120
Which is also true
P(n) : 2n < n! is true for n > 3
P(n) : 2n < n! is true for n > 3

#### Question 4:

For each nN, 102n – 1 + 1 is divisible by _____________.

For each n ∈ N,
Let P(n) : 102n–1 + 1
for n = 1
L.H.S = 102(1)–1 + 1
= 101 + 1
= 10 + 1
= 11
i.e P(1) = 11
Assume P(n) is true

for = 2,

Similarly, assume that P(k) is divisible by 11.

#### Question 5:

If P(n) : n! > 2n – 1, nN, then P(n) is true for all n > _____________.

P(n) : n! > 2n – 1n ∈ N

for n = 1,
P(1) : 1! > 21–1
i.e 1 > 2° = 1
i.e 1 > 1
which is false a statement

for n = 2
P(2) : 2! > 22–1
i.e 2 > 21
i.e 2 > 2
which is again a false statement.

for n = 3
P(3) : 3! > 23–1
i.e 6 > 22 = 4
i.e 6 > 4 which is true

for n = 4
P(4) : 4! > 24–1
i.e 24 > 2= 8 which is true

Hence, P(n) : n! > 2n – 1 is true
for n > 2

#### Question 6:

If P(n) : n3n is divisible by 6, nN, then P(n) is true for all n ≥ _____________.

P(n) : n3n is divisible by 6; nN

for n = 1,
P(1) : (1)3 – 1 = 0 which is divisible by 6

for n = 2,
P(2) : (2)3 – 2
= 8 – 2 = 6 which is divisible by 6

for n = 3,
P(3) : (3)3 – 3
= 27 – 3
= 24;  which is divisible by 6

Hence, P(n) is true for n ≥ 2

#### Question 7:

If P(n) : n2 < 2n, nN, then P(n) is true for all n ≥ _____________.

P(n) : n2 < 2n, nN

for n = 1,
P(1) : 1 < 2  which is true statement

for n = 2,
P(2) : 22 < 22   which is not true/false statement

for n = 3,
P(3) : (3)2 < 23
i.e 9 < 8 which is a false statement

for n = 4,
P(4) : (4)2 < 24
i.e 16 < 4 × 4 = 16
which is a false statement

for n = 5,
P(5) : (5)2 < 25
i.e 25 < 32
which is a true statement

for n = 6,
P(6) : (36) < 26
i.e 36 < 4 × 4 × 4 = 64
Which is a true statement

for n = 7,  P(7) : 49 < 27 = 12 which is again true statement
Hence, P(n) : n2 < 2n is true for n ≥ 5

#### Question 8:

If   then P(n) is true for all n ≥ _____________.

for n = 1,

which is a false statement.

for n = 2,

which is true

for n = 3

which is again true
Hence, .

#### Question 1:

State the first principle of mathematical induction.

Let P(n) be a given statement involving the natural number n such that

(i) The statement is true for n = 1, i.e., P(1) is true (or true for any fixed natural number). This step is known as the Basis step.

(ii) If the statement (called Induction hypothesis) is true for n = k (where k is a particular but arbitrary natural number), then the statement is also true for n = k + 1,

i.e, truth of P(k) implies the truth of P(k + 1). This step is known as the Induction (or Inductive) step.

Then P(n) is true for all natural numbers n.

Note: The first principle of mathematical induction states that if the basis step and the inductive step are proven, then P(nis true for all natural numbers.

#### Question 2:

Write the set of value of n for which the statement P(n): 2n < n! is true.

As, n! > 2n when n > 3.

So, the set of value of n for which the statement P(n): 2n < n! is true = {n $\in$ N: n > 3}.

#### Question 3:

State the second principle of mathematical induction.

Second principle of mathematical induction:

Let P(n) be a given statement involving the natural number n such that

(i) The statement is true for n = 1, i.e., P(1) is true (or true for any fixed natural number). This step is known as the Basis step.

(ii) If the statement (called Induction hypothesis) is true for 1 $\le$ n $\le$ k (where k is a particular but arbitrary natural number), then the statement is also true for n = k + 1,

i.e, truth of P(k) implies the truth of P(k + 1). This step is known as the Induction (or Inductive) step.

Then P(n) is true for all natural numbers n.

Note: The second principle of mathematical induction is completely equivalent to the first principle of mathematical induction which states that if the basis step and the inductive step are proven, then P(n) is true for all natural numbers.

But the only difference is in the inductive hypothesis step
that we assume not only that the statement holds for n = k but also that it is true for all $\le$ n $\le$ k.

Also, the base can be other natural number as well apart 1 in both the principles.



#### Question 4:

If P(n): $2×{4}^{2n+1}+{3}^{3n+1}$ is divisible by $\lambda$ for all n $\in$ N is true, then find the value of $\lambda$.

#### Question 1:

If P (n) is the statement "n(n + 1) is even", then what is P(3)?

We have:
P(n): n(n + 1) is even.
Now,
P(3) = 3(3 + 1) = 12       (Even)
Therefore, P(3) is even.

#### Question 2:

If P (n) is the statement "n3 + n is divisible by 3", prove that P (3) is true but P (4) is not true.

We have:

#### Question 3:

If P (n) is the statement "2n ≥ 3n" and if P (r) is true, prove that P (r + 1) is true.

We have:

#### Question 4:

If P (n) is the statement "n2 + n is even", and if P (r) is true, then P (r + 1) is true.

#### Question 5:

Given an example of a statement P (n) such that it is true for all n ∈ N.

Proved:
)

#### Question 6:

If P (n) is the statement "n2n + 41 is prime", prove that P (1), P (2) and P (3) are true. Prove also that P (41) is not true.

#### Question 7:

Give an example of a statement P(n) which is true for all n ≥ 4 but P(1), P(2) and P(3) are not true. Justify your answer.

Let P(n) be the statement 3n < n!.

For n = 1,

3n = 3 × 1 = 3

n! = 1! = 1

Now, 3 > 1

So, P(1) is not true.

For n = 2,

3n = 3 × 2 = 6

n! = 2! = 2

Now, 6 > 2

So, P(2) is not true.

For n = 3,

3n = 3 × 3 = 9

n! = 3! = 6

Now, 9 > 6

So, P(3) is not true.

For n = 4,

3n = 3 × 4 = 12

n! = 4! = 24

Now, 12 < 24

So, P(4) is true.

For n = 5,

3n = 3 × 5 = 15

n! = 5! = 120

Now, 15 < 120

So, P(5) is true.

Similarly, it can be verified that 3n < n! for n = 6, 7, 8, ... .

Thus, the statement P(n) : 3n < n! is true for all n ≥ 4 but P(1), P(2) and P(3) are not true.

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