We have provided Class 11 Maths MCQ Questions with Answers to help students understand the concept very well. Class 11 Maths MCQ Questions of Principle of Mathematical Induction with Answers were prepared based on the latest exam pattern.

MCQ Questions with Answers provide here with detailed solutions so that Students can easily understand the logic behind each answer. It will help students to attempt the exam with confidence. You also get ideas about the type of questions and methods to answer in your Class 11th examination. Practice Class 11 Maths MCQ Questions of Principle of Mathematical Induction given below.

Practice MCQ Questions for class 11 Maths Chapter-Wise

1. 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

2. 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

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

(a) n(n + 1)
(b) n/(n + 1)
(c) 2n/(n + 1)
(d) 3n/(n + 1)

4. 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

5. {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

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

(a) 3
(b) 4
(c) 5
(d) 7

7. The nth terms of the series 3 + 7 + 13 + 21 +………. is

(a) 4n – 1
(b) n² + n + 1
(c) none of these
(d) n + 2

8. Find the number of shots arranged in a complete pyramid the base of which is an equilateral triangle, each side containing n shots.

(a) n(n+1)(n+2)/3
(b) n(n+1)(n+2)/6
(c) n(n+2)/6
(d) (n+1)(n+2)/6

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

(a) 3
(b) 4
(c) 5
(d) 7

10. For all n ∈ N, 3 × 5²ⁿ⁺¹ + 2³ⁿ⁺¹ is divisible by

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

11. For all n ∈ N, 7²ⁿ − 48n −1 is divisible by :

(a) 25
(b) 2304
(c) 1234
(d) 26

12. Let P(n) be the statement 2ⁿ <n! where n is a natural number, then P(n) is true for:

(a) all n
(b) all n>2
(c) all n>3
(d) all n<3

13. Let T(k) be the statement 1+3+5+....+(2k−1) = k²+10. Which of the following is correct?

(a) T(1) is true
(b) T(k) is true ⇒ T(k + 1) is true
(c) T(n) is true for all n ∈ N
(d) All above are correct

14. Let P(n)  :n²+n+1 is an even integer. If P(k) is assumed true ⇒ P(k+1) is true. Therefore, P(n) is

(a) true for n>1,n∈N
(b) true for all n∈N
(c) true for n>2,n∈N
(d) none of these

15. If x ⁿ⁻¹ is divisible by x−k, then the least positive integral value of k is

(a) 1
(b) 2
(c) 3
(d) 4

16. if n is a +ve integer, then 2.4²ⁿ⁺¹ + 3³ⁿ⁺¹is divisible by

(a) 2
(b) 7
(c) 11
(d) 27

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

(a) for all n ∈ N
(b) for all n > 5
(c) for all n ≥ 5
(d) for all n < 5

18.The greatest positive integer, which divides n(n +1)(n + 2)(n + 3) for all n ∈ N , is

(a) 2
(b) 6
(c) 24
(d) 120

19. n(n + 1) (n + 5) is a multiple of

(a) 3
(b) 8
(c) 5
(d) 7

20. For every positive integer n, 7n – 3n is divisible by

(a) 3
(b) 4
(c) 5
(d) 7

Answer:

1. Answer: (d) {n(n + 1)/2}²

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

Sum = {n(n + 1)/2}²

2. Answer: (b) a + b

Explanation: Given number = an + bn

Let n = 1, 3, 5, ……..

a_{n +} b_n = a + b

a_n + b_n = 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)

3. Answer: (b) n/(n + 1)

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

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

Putting n = 1 in the given statement, we get

LHS = 1/(1 ∙ 2) = and RHS = 1/(1 + 1) = 1/2.

LHS = RHS.

Thus, P(1) is true.

Let P(k) be true. Then,

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

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

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

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

= {k(k + 2) + 1}/{(k + 1)²/[(k + 1)k + 2)] using …(ii)

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

= {(k + 1)² }/{(k + 1)(k + 2)}

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

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

= (k + 1)/(k + 1 + 1)

⇒ 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.

4. Answer: (d) n(n + 1)(2n + 1)/6

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

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

5. Answer: (a) 1/(n + 1) for all n ∈ N.

Explanation: 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)} = 1/2 and RHS = 1/(1 + 1) = 1/2.

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.

6. Answer: (c) 5

Explanation: Given, 7ⁿ – 2ⁿ

Let n = 1

7ⁿ – 2ⁿ = 7¹ – 2¹ 

= 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.

7. Answer: (b) n² + n + 1

Explanation: Let S = 3 + 7 + 13 + 21 +……….a_n-1 + an …………1

and S = 3 + 7 + 13 + 21 +……….a_n-1 + an …………2

Subtract equation 1 and 2, we get

S – S = 3 + (7 + 13 + 21 +……….a_n-1 + a_n) – (3 + 7 + 13 + 21 +……….a_n-1 + a_n)

⇒ 0 = 3 + (7 – 3) + (13 – 7) + (21 – 13) + ……….+ (a_n – a_n-1) – a_n

⇒ 0 = 3 + {4 + 6 + 8 + ……(n-1)terms} – an

⇒ a_n = 3 + {4 + 6 + 8 + ……(n-1)terms}

⇒ a_n = 3 + (n – 1)/2 × {2 ×4 + (n – 1 – 1)2}

⇒ a_n = 3 + (n – 1)/2 × {8 + (n – 2)2}

⇒ a_n = 3 + (n – 1) × {4 + n – 2}

⇒ a_n = 3 + (n – 1) × (n + 2)

⇒ a_n = 3 + n²+ n – 2

⇒ an = n² + n + 1

So, the nth term is n² + n + 1

8. Answer: (d) (n+1)(n+2)/6

Explanation: Let each side of the base contains n shots,

then the number of shots in the lowest layer = n + (n – 1) + (n – 2) + ………..+ 1

= n(n + 1)/2

= (n  + n)/2

Now, write (n – 1), (n – 2), ….. for n, then we obtain the number of shots in 2nd, 3rd…layers

So, Total shots = ∑(n² + n)/2

= (1/2)×{∑n² + ∑n}

= (1/2)×{n(n+1)(2n+1)/6 + n(n+1)/2}

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

9. Answer: (c) 5

Explanation: Given, 7ⁿ – 2ⁿ

Let n = 1

7n – 2n 

= 7¹ – 2¹ 

= 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

10. Answer: (b) 17

Explanation: Given, 3 × 5²ⁿ⁺¹ + 2³ⁿ⁺¹

Let n = 1,

3 × 5^2×1+1 + 2^3×1+1 

= 3 × 5²⁺¹ + 2³⁺¹ 

= 3 × 5³ + 2⁴ 

= 3 × 125 + 16 

= 375 + 16 

= 391

Which is divisible by 17

Let n = 2,

3 × 5^2×2+1 + 2^3×2+1 

= 3 × 5⁴⁺¹ + 2⁶⁺¹ 

= 3 × 5⁵ + 2⁷ 

= 3 × 3125 + 128 

= 9375 + 128

= 9503

Which is divisible by 17

Hence, For all n ∈ N, 3 × 5²ⁿ⁺¹ + 2³ⁿ⁺¹ is divisible by 17

11. Answer: (b) 2304

Explanation: Given number = 72n − 48n − 1

Let n = 1, 2 ,3, 4, ……..

7²ⁿ − 48n − 1 = 72 − 48 − 1 

= 49 – 48 – 1 

= 49 – 49 

= 0

7²ⁿ − 48n − 1 

= 7⁴ − 48 × 2 − 1 

= 2401 – 96 – 1 

= 2401 – 97 

= 2304

7²ⁿ − 48n − 1 

= 7⁶ − 48 × 3 − 1

= 117649 – 144 – 1 

= 117649 – 145 

= 117504 

= 2304 × 51

Since, all these numbers are divisible by 2304 for n = 1, 2, 3,…..

So, the given number is divisible by 2304

12. Answer: (c) all n>3

Explanation: We have,P(n) be the statement 2ⁿ <n!

Where n is a natural number

Then,

Put n=1,2,3....

So, P(1) be the statement of 2¹<1! = 2<1 (It is wrong)

P(2) be the statement of 2² < 2! = 4<2 (It is wrong)

P(3)  be the statement of  <3!=8<6(It is wrong)

But,

P(4)  be the statement of 2⁴ <4! =16<24(Itisright)

Similarly,

P(5), P(6)....... So, all number n is a natural number.

13. Answer: (b) T(k) is true ⇒ T(k + 1) is true

Explanation: When k = 1, LHS = 1 but RHS = 1 + 10 = 11

∴ T(1) is not true

Let T(k) is true.

i.e., 1+3+5+.....+(2k−1) = k²+10

Now, 1+ 3+ 5 + ..... + (2k -1) + (2k +1)

= k²+10+2k+1 = (k+1)²+10

∴ T(k +1) is true.

i.e., T(k) is true ⇒ T (k +1) is true.

But T(n) is not true for all n ∈ N , as T(1) is not true.

14. Answer: (d) none of these    

Explanation: Given, P(n)  :n²+n+1 is an even integer.

For n=1,P(1)=3, which is not an even integer.

∴P(1) is not true.

Also, n²+n+1 = n(n+1)+1 is always an odd integer. (principle of induction is not applicable).

15. Answer: (a) 1

Explanation: x⁻¹ is always a factor of xⁿ⁻¹.

⇒k =1.

16. Answer: (c) 11

Explanation: Let P(n) =2.4²ⁿ⁺¹ + 3³ⁿ⁺¹

Then P(1)=2.4³+3⁴ =209, which is divisible by 11 but not divisible by 2, 7 or 27.

Further, let P(k)=2.4^2k+1 + 3^3k+1 is divisible by 11, i.e., 2.4^2k+1+ 3^3k+1=11q for some integer q.

Now P(k+1)=2.4^2k+3 + 3^3k+4

=2.4^2k+1.4² + 3^3k+1.3³

=16.2.4^2k+1+27.3^3k+1

=16.2.2^2k+1+(16+11).3^3k+1

=16[2.4^2k+1+3^3k+1]+11.3^3k+1

=16.11q+11.3^3k+1

=11(16q+3^3k+1)

=11m

where m=16q+33k+1 is another integer.

∴p(k+1) is divisible by 11.

∴P(n)=2.4²ⁿ⁺¹+3³ⁿ⁺¹ is divisible by 11

for all n∈N.

17. Answer: (c) for all n ≥ 5

Explanation: Since P(5) is true and P(k + 1) is true, whenever P (k) is true.

18. Answer: (c) 24

Explanation: The product of r consecutive integers is divisible by r ! . Thus n (n+ 1 ) (n + 2) (n + 3) is divisible by 4 ! = 24.

19. Answer: (a) 3

Explanation: Let P(n) = n(n + 1)(n + 5)

Substituting n = 1, 2, 3,….

P(1) = 1(1 + 1)(1 + 5) = 2(6) = 12; multiple of 2, 3, 4, 6

P(2) = 2(2 + 1)(2 + 5) = 2(3)(7) = 42; multiple of 2, 3, 6, 7

P(3) = 3(3 + 1)(3 + 5) = 3(4)(8) = 96; multiple of 2, 3, 4, 6, 8, 12

20. Answer: (b) 4

Explanation: Let P(n) = 7ⁿ – 3ⁿ

Substituting n = 1, 2, 3,…

P(1) = 71 – 31 = 7 – 3 = 4

P(2) = 72 – 32 = 49 – 9 = 40

P(3) = 73 – 33 = 343 – 27 = 316

Thus, for every positive integer n, 7n – 3n is divisible by 4.

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