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

Students are advised to solve the Principle of Mathematical Induction Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Principle of Mathematical Induction Class 11 with answers will boost your confidence thereby helping you score well in the exam.

Explore numerous MCQ Questions of Principle of Mathematical Induction Class 11 with answers provided with detailed solutions by looking below.

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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