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## Principle of Mathematical Induction Class 11 MCQs Questions with Answers

Solving the MCQ Questions of Principle of Mathematical Induction Class 11 with answers can help you understand the concepts better.

### The Principle of Mathematical Induction

Suppose there is a given statement P(n) involving the natural number n such that

(i) The statement is true for n = 1, i.e., P(1) is true, and

(ii) If the statement is true for n = k (where k is some positive integer), then the statement is also true for n = k + 1, i.e., truth of P(k) implies the truth of P (k + 1).

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

Property (i) is simply a statement of fact. There may be situations when a statement is true for all n ≥ 4. In this case, step 1 will start from n = 4 and we shall verify the result for n = 4, i.e., P(4).

Property (ii) is a conditional property. It does not assert that the given statement is true for n = k, but only that if it is true for n = k, then it is also true for n = k +1. So, to prove that the property holds, only prove that conditional proposition:

If the statement is true for n = k, then it is also true for n = k + 1.

This is sometimes referred to as the inductive step. The assumption that the given statement is true for n = k in this inductive step is called the inductive hypothesis.

Question 1.

For all n∈N, 3n^{5} + 5n³ + 7n is divisible by

(a) 5

(b) 15

(c) 10

(d) 3

## Answer

Answer: (b) 15

Given number = 3n^{5} + 5n² + 7n

Let n = 1, 2, 3, 4, ……..

3n^{5} + 5n³ + 7n = 3 × 1² + 5 × 1³ + 7 × 1 = 3 + 5 + 7 = 15

3n^{5} + 5n³ + 7n = 3 × 2^{5} + 5 × 2³ + 7 × 2 = 3 × 32 + 5 × 8 + 7 × 2 = 96 + 40 + 14 = 150 = 15 × 10

3n^{5} + 5n³ + 7n = 3 × 3^{5} + 5 × 3³ + 7 × 3 = 3 × 243 + 5 × 27 + 7 × 3 = 729 + 135 + 21 = 885 = 15 × 59

Since, all these numbers are divisible by 15 for n = 1, 2, 3, …..

So, the given number is divisible by 15

Question 2.

{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. {1 – 1/(n + 1)} =

(a) 1/(n + 1) for all n ∈ N.

(b) 1/(n + 1) for all n ∈ R

(c) n/(n + 1) for all n ∈ N.

(d) n/(n + 1) for all n ∈ R

## Answer

Answer: (a) 1/(n + 1) for all n ∈ N.

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

P(n): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. {1 – 1/(n + 1)} = 1/(n + 1).

When n = 1, LHS = {1 – (1/2)} = ½ and RHS = 1/(1 + 1) = ½.

Therefore LHS = RHS.

Thus, P(1) is true.

Let P(k) be true. Then,

P(k): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 1)

Now, [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] ∙ [1 – {1/(k + 2)}]

= [1/(k + 1)] ∙ [{(k + 2 ) – 1}/(k + 2)}]

= [1/(k + 1)] ∙ [(k + 1)/(k + 2)]

= 1/(k + 2)

Therefore p(k + 1): [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 2)

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 3.

For all n ∈ N, 3^{2n} + 7 is divisible by

(a) non of these

(b) 3

(c) 11

(d) 8

## Answer

Answer: (d) 8

Given number = 32n + 7

Let n = 1, 2, 3, 4, ……..

3^{2n} + 7 = 3² + 7 = 9 + 7 = 16

3^{2n} + 7 = 3^{4} + 7 = 81 + 7 = 88

3^{2n} + 7 = 3^{6} + 7 = 729 + 7 = 736

Since, all these numbers are divisible by 8 for n = 1, 2, 3, …..

So, the given number is divisible by 8

Question 4.

The sum of the series 1 + 2 + 3 + 4 + 5 + ………..n is

(a) n(n + 1)

(b) (n + 1)/2

(c) n/2

(d) n(n + 1)/2

## Answer

Answer: (d) n(n + 1)/2

Given, series is series 1 + 2 + 3 + 4 + 5 + ………..n

Sum = n(n + 1)/2

Question 5.

The sum of the series 1² + 2² + 3² + ……….. n² is

(a) n(n + 1) (2n + 1)

(b) n(n + 1) (2n + 1)/2

(c) n(n + 1) (2n + 1)/3

(d) n(n + 1) (2n + 1)/6

## Answer

Answer: (d) n(n + 1) (2n + 1)/6

Given, series is 1² + 2² + 3² + ……….. n²

Sum = n(n + 1)(2n + 1)/6

Question 6.

For all positive integers n, the number n(n² − 1) is divisible by:

(a) 36

(b) 24

(c) 6

(d) 16

## Answer

Answer: (c) 6

Given,

number = n(n² − 1)

Let n = 1, 2, 3, 4….

n(n² – 1) = 1(1 – 1) = 0

n(n² – 1) = 2(4 – 1) = 2 × 3 = 6

n(n² – 1) = 3(9 – 1) = 3 × 8 = 24

n(n² – 1) = 4(16 – 1) = 4 × 15 = 60

Since all these numbers are divisible by 6 for n = 1, 2, 3,……..

So, the given number is divisible 6

Question 7.

If n is an odd positive integer, then aⁿ + bⁿ is divisible by :

(a) a² + b²

(b) a + b

(c) a – b

(d) none of these

## Answer

Answer: (b) a + b

Given number = aⁿ + bⁿ

Let n = 1, 3, 5, ……..

aⁿ + bⁿ = a + b

aⁿ + bⁿ = a³ + b³ = (a + b) × (a² + b² + ab) and so on.

Since, all these numbers are divisible by (a + b) for n = 1, 3, 5,…..

So, the given number is divisible by (a + b)

Question 8.

n(n + 1) (n + 5) is a multiple of ____ for all n ∈ N

(a) 2

(b) 3

(c) 5

(d) 7

## Answer

Answer: (b) 3

Let P(n): n(n + 1)(n + 5) is a multiple of 3.

For n = 1, the given expression becomes (1 × 2 × 6) = 12, which is a multiple of 3.

So, the given statement is true for n = 1, i.e. P(1) is true.

Let P(k) be true. Then,

P(k): k(k + 1)(k + 5) is a multiple of 3

⇒ K(k + 1) (k + 5) = 3m for some natural number m, …… (i)

Now, (k + 1) (k + 2) (k + 6) = (k + 1) (k + 2)k + 6(k + 1) (k + 2)

= k(k + 1) (k + 2) + 6(k + 1) (k + 2)

= k(k + 1) (k + 5 – 3) + 6(k + 1) (k + 2)

= k(k + 1) (k + 5) – 3k(k + 1) + 6(k + 1) (k + 2)

= k(k + 1) (k + 5) + 3(k + 1) (k +4) [on simplification]

= 3m + 3(k + 1 ) (k + 4) [using (i)]

= 3[m + (k + 1) (k + 4)], which is a multiple of 3

⇒ P(k + 1) (k + 1 ) (k + 2) (k + 6) is a multiple of 3

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 9.

For any natural number n, 7ⁿ – 2ⁿ is divisible by

(a) 3

(b) 4

(c) 5

(d) 7

## Answer

Answer: (c) 5

Given, 7ⁿ – 2ⁿ

Let n = 1

7ⁿ – 2ⁿ = 7^{1} – 2^{1} = 7 – 2 = 5

which is divisible by 5

Let n = 2

7ⁿ – 2ⁿ = 72 – 22 = 49 – 4 = 45

which is divisible by 5

Let n = 3

7ⁿ – 2ⁿ = 7³ – 2³ = 343 – 8 = 335

which is divisible by 5

Hence, for any natural number n, 7ⁿ – 2ⁿ is divisible by 5

Question 10.

The sum of the series 1³ + 2³ + 3³ + ………..n³ is

(a) {(n + 1)/2}²

(b) {n/2}²

(c) n(n + 1)/2

(d) {n(n + 1)/2}²

## Answer

Answer: (d) {n(n + 1)/2}²

Given, series is 1³ + 2³ + 3³ + ……….. n³

Sum = {n(n + 1)/2}²

Question 11.

(1² + 2² + …… + n²) _____ for all values of n ∈ N

(a) = n³/3

(b) < n³/3

(c) > n³/3

(d) None of these

## Answer

Answer: (c) > n³/3

Let P(n): (1² + 2² + ….. + n²) > n³/3.

When = 1, LHS = 1² = 1 and RHS = 1³/3 = 1/3.

Since 1 > 1/3, it follows that P(1) is true.

Let P(k) be true. Then,

P(k): (1² + 2² + ….. + k² ) > k³/3 …. (i)

Now,

1² + 2² + ….. + k²

+ (k + 1)²

= {1² + 2² + ….. + k² + (k + 1)²

> k³/3 + (k + 1)³ [using (i)]

= 1/3 ∙ (k³ + 3 + (k + 1)²) = 1/3 ∙ {k² + 3k² + 6k + 3}

= 1/3[k³ + 1 + 3k(k + 1) + (3k + 2)]

= 1/3 ∙ [(k + 1)³ + (3k + 2)]

> 1/3(k + 1)³

P(k + 1):

1² + 2² + ….. + k² + (k + 1)²

> 1/3 ∙ (k + 1)³

P(k + 1) is true, whenever P(k) is true.

Thus P(1) is true and P(k + 1) is true whenever p(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 12.

{1/(3 ∙ 5)} + {1/(5 ∙ 7)} + {1/(7 ∙ 9)} + ……. + 1/{(2n + 1) (2n + 3)} =

(a) n/(2n + 3)

(b) n/{2(2n + 3)}

(c) n/{3(2n + 3)}

(d) n/{4(2n + 3)}

## Answer

Answer: (c) n/{3(2n + 3)}

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

P(n): {1/(3 ∙ 5) + 1/(5 ∙ 7) + 1/(7 ∙ 9) + ……. + 1/{(2n + 1)(2n + 3)} = n/{3(2n + 3).

Putting n = 1 in the given statement, we get

and LHS = 1/(3 ∙ 5) = 1/15 and RHS = 1/{3(2 × 1 + 3)} = 1/15.

LHS = RHS

Thus, P(1) is true.

Let P(k) be true. Then,

P(k): {1/(3 ∙ 5) + 1/(5 ∙ 7) + 1/(7 ∙ 9) + …….. + 1/{(2k + 1)(2k + 3)} = k/{3(2k + 3)} ….. (i)

Now, 1/(3 ∙ 5) + 1/(5 ∙ 7) + ..…… + 1/[(2k + 1)(2k + 3)] + 1/[{2(k + 1) + 1}2(k + 1) + 3

= {1/(3 ∙ 5) + 1/(5 ∙ 7) + ……. + [1/(2k + 1)(2k + 3)]} + 1/{(2k + 3)(2k + 5)}

= k/[3(2k + 3)] + 1/[2k + 3)(2k + 5)] [using (i)]

= {k(2k + 5) + 3}/{3(2k + 3)(2k + 5)}

= (2k² + 5k + 3)/[3(2k + 3)(2k + 5)]

= {(k + 1)(2k + 3)}/{3(2k + 3)(2k + 5)}

= (k + 1)/{3(2k + 5)}

= (k + 1)/[3{2(k + 1) + 3}]

= P(k + 1): 1/(3 ∙ 5) + 1/(5 ∙ 7) + …….. + 1/[2k + 1)(2k + 3)] + 1/[{2(k + 1) + 1}{2(k + 1) + 3}]

= (k + 1)/{3{2(k + 1) + 3}]

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for n ∈ N.

Question 13.

If n is an odd positive integer, then aⁿ + bⁿ is divisible by :

(a) a² + b²

(b) a + b

(c) a – b

(d) none of these

## Answer

Answer: (b) a + b

Given number = aⁿ + bⁿ

Let n = 1, 3, 5, ……..

aⁿ + bⁿ = a + b

aⁿ + bⁿ = a³ + b³ = (a + b) × (a² + b² + ab) and so on.

Since, all these numbers are divisible by (a + b) for n = 1, 3, 5,…..

So, the given number is divisible by (a + b)

Question 14.

(2 ∙ 7^{N} + 3 ∙ 5^{N} – 5) is divisible by ……….. for all N ∈ N

(a) 6

(b) 12

(c) 18

(d) 24

## Answer

Answer: (d) 24

Let P(n): (2 ∙ 7ⁿ + 3 ∙ 5ⁿ – 5) is divisible by 24.

For n = 1, the given expression becomes (2 ∙ 7^{1} + 3 ∙ 5^{1} – 5) = 24, which is clearly divisible by 24.

So, the given statement is true for n = 1, i.e., P(1) is true.

Let P(k) be true. Then,

P(k): (2 ∙ 7ⁿ + 3 ∙ 5ⁿ – 5) is divisible by 24.

⇒ (2 ∙ 7ⁿ + 3 ∙ 5ⁿ – 5) = 24m, for m = N

Now, (2 ∙ 7ⁿ + 3 ∙ 5ⁿ – 5)

= (2 ∙ 7^{k} ∙ 7 + 3 ∙ 5^{k} ∙ 5 – 5)

= 7(2 ∙ 7^{k} + 3 ∙ 5^{k} – 5) = 6 ∙ 5^{k} + 30

= (7 × 24m) – 6(5^{k} – 5)

= (24 × 7m) – 6 × 4p, where (5^{k} – 5) = 5(5^{k-1} – 1) = 4p

[Since (5^{k-1} – 1) is divisible by (5 – 1)]

= 24 × (7m – p)

= 24r, where r = (7m – p) ∈ N

⇒ P (k + 1): (2 ∙ 7^{k} + 13 ∙ 5^{k} + 1 – 5) is divisible by 24.

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 15.

For all n∈N, 5^{2n} − 1 is divisible by

(a) 26

(b) 24

(c) 11

(d) 25

## Answer

Answer: (b) 24

Given number = 5^{2n} − 1

Let n = 1, 2, 3, 4, ……..

5^{2n} − 1 = 5² − 1 = 25 – 1 = 24

5^{2n} − 1 = 5^{4} – 1 = 625 – 1 = 624 = 24 × 26

5^{2n} − 1 = 5^{6} – 1 = 15625 – 1 = 15624 = 651 × 24

Since, all these numbers are divisible by 24 for n = 1, 2, 3, …..

So, the given number is divisible by 24

Question 16.

1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ….. + n(n + 1) =

(a) n(n + 1)(n + 2)

(b) {n(n + 1)(n + 2)}/2

(c) {n(n + 1)(n + 2)}/3

(d) {n(n + 1)(n + 2)}/4

## Answer

Answer: (c) {n(n + 1)(n + 2)}/3

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

P(n): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ….. + n(n + 1) = (1/3){n(n + 1) (n + 2)}

Thus, the given statement is true for n = 1, i.e., P(1) is true.

Let P(k) be true. Then,

P(k): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ….. + k(k + 1) = (1/3){k(k + 1) (k + 2)}.

Now, 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 +…+ k(k + 1) + (k + 1) (k + 2)

= (1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ……. + k(k + 1)) + (k + 1) (k + 2)

= (1/3) k(k + 1) (k + 2) + (k + 1)(k + 2) [using (i)]

= (1/3) [k(k + 1) (k + 2) + 3(k + 1)(k + 2)

= (1/3){(k + 1) (k + 2)(k + 3)}

⇒ P(k + 1): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 +……+ (k + 1) (k + 2)

= (1/3){k + 1 )(k + 2) (k +3)}

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1)is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all values of ∈ N.

Question 17.

1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + …….. + 1/{n(n + 1)(n + 2)} =

(a) {n(n + 3)}/{4(n + 1)(n + 2)}

(b) (n + 3)/{4(n + 1)(n + 2)}

(c) n/{4(n + 1)(n + 2)}

(d) None of these

## Answer

Answer: (a) {n(n + 3)}/{4(n + 1)(n + 2)}

Let P (n): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……. + 1/{n(n + 1)(n + 2)} = {n(n + 3)}/{4(n + 1)(n + 2)}

Putting n = 1 in the given statement, we get

LHS = 1/(1 ∙ 2 ∙ 3) = 1/6 and RHS = {1 × (1 + 3)}/[4 × (1 + 1)(1 + 2)] = ( 1 × 4)/(4 × 2 × 3) = 1/6.

Therefore LHS = RHS.

Thus, the given statement is true for n = 1, i.e., P(1) is true.

Let P(k) be true. Then,

P(k): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……… + 1/{k(k + 1) (k + 2)} = {k(k + 3)}/{4(k + 1) (k + 2)}. …….(i)

Now, 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………….. + 1/{k(k + 1) (k + 2)} + 1/{(k + 1) (k + 2) (k + 3)}

= [1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………..…. + 1/{ k(k + 1) (k + 2}] + 1/{(k + 1)(k + 2) (k + 3)}

= [{k(k + 3)}/{4(k + 1)(k + 2)} + 1/{(k + 1)(k + 2)(k + 3)}] [using(i)]

= {k(k + 3)² + 4}/{4(k + 1)(k + 2) (k + 3)}

= (k³ + 6k² + 9k + 4)/{4(k + 1) (k + 2) (k + 3)}

= {(k + 1) (k + 1) (k + 4)}/{4 (k + 1) (k + 2) (k + 3)}

= {(k + 1) (k + 4)}/{4(k + 2) (k + 3)

⇒ P(k + 1): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……….….. + 1/{(k + 1) (k + 2) (k + 3)}

= {(k + 1) (k + 2)}/{4(k + 2) (k + 3)}

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 18.

For any natural number n, 7ⁿ – 2ⁿ is divisible by

(a) 3

(b) 4

(c) 5

(d) 7

## Answer

Answer: (c) 5

Given, 7ⁿ – 2ⁿ

Let n = 1

7ⁿ – 2ⁿ = 7^{1} – 2^{1} = 7 – 2 = 5

which is divisible by 5

Let n = 2

7ⁿ – 2ⁿ = 7² – 2² = 49 – 4 = 45

which is divisible by 5

Let n = 3

7ⁿ – 2ⁿ = 7³ – 2³ = 343 – 8 = 335

which is divisible by 5

Hence, for any natural number n, 7ⁿ – 2ⁿ is divisible by 5

Question 19.

The sum of n terms of the series 1² + 3² + 5² +……… is

(a) n(4n² – 1)/3

(b) n²(2n² + 1)/6

(c) none of these.

(d) n²(n² + 1)/3

## Answer

Answer: (a) n(4n² – 1)/3

Let S = 1² + 3² + 5² +………(2n – 1)²

⇒ S = {1² + 2² + 3² + 4² ………(2n – 1)² + (2n)²} – {2² + 4² + 6² +………+ (2n)²}

⇒ S = {2n × (2n + 1) × (4n + 1)}/6 – {4n × (n + 1) × (2n + 1)}/6

⇒ S = n(4n² – 1)/3

Question 20.

For all n ∈ N, 3n^{5} + 5n³ + 7n is divisible by:

(a) 5

(b) 15

(c) 10

(d) 3

## Answer

Answer: (b) 15

Given number = 3n^{5} + 5n³ + 7n

Let n = 1, 2, 3, 4, ……..

3n^{5} + 5n³ + 7n = 3 × 1² + 5 × 1³ + 7 × 1 = 3 + 5 + 7 = 15

3n^{5} + 5n³ + 7n = 3 × 2^{5} + 5 × 2³ + 7 × 2 = 3 × 32 + 5 × 8 + 7 × 2 = 96 + 40 + 14 = 150 = 15 × 10

3n^{5} + 5n³ + 7n = 3 × 3^{5} + 5 × 3³ + 7 × 3 = 3 × 243 + 5 × 27 + 7 × 3 = 729 + 135 + 21 = 885 = 15 × 59

Since, all these numbers are divisible by 15 for n = 1, 2, 3, …..

So, the given number is divisible by 15

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