 Check the below NCERT MCQ Questions for Class 11 Maths Chapter 4 Principle of Mathematical Induction with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have provided Principle of Mathematical Induction Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

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

Mathematical Induction MCQ Question 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

Hint:
Given, series is 1³ + 2³ + 3³ + ……….. n³
Sum = {n(n + 1)/2}²

MCQ On Mathematical Induction Question 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

Hint:
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)

MCQ Questions On Mathematical Induction Question 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)

Hint:
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.

Mathematical Induction MCQs Pdf Question 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

Answer: (d) n(n + 1)(2n + 1)/6
Hint:
Given, series is 1² + 2² + 3² + ………..n²
Sum = n(n + 1)(2n + 1)/6

Class 11 Maths Chapter 4 MCQ With Answers Question 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

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

Mathematical Induction MCQs Question 6.
For any natural number n, 7n – 2n is divisible by
(a) 3
(b) 4
(c) 5
(d) 7

Hint:
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

Principle Of Mathematical Induction Class 11 MCQs Question 7.
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: (a) {n(n + 3)}/{4(n + 1)(n + 2)}
Hint:
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.

MCQs On Mathematical Induction Question 8.
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

Answer: (b) n² + n + 1
Hint:
Let S = 3 + 7 + 13 + 21 +……….an-1 + an …………1
and S = 3 + 7 + 13 + 21 +……….an-1 + an …………2
Subtract equation 1 and 2, we get
S – S = 3 + (7 + 13 + 21 +……….an-1 + an) – (3 + 7 + 13 + 21 +……….an-1 + an)
⇒ 0 = 3 + (7 – 3) + (13 – 7) + (21 – 13) + ……….+ (an – an-1) – an
⇒ 0 = 3 + {4 + 6 + 8 + ……(n-1)terms} – an
⇒ an = 3 + {4 + 6 + 8 + ……(n-1)terms}
⇒ an = 3 + (n – 1)/2 × {2 ×4 + (n – 1 – 1)2}
⇒ an = 3 + (n – 1)/2 × {8 + (n – 2)2}
⇒ an = 3 + (n – 1) × {4 + n – 2}
⇒ an = 3 + (n – 1) × (n + 2)
⇒ an = 3 + n² + n – 2
⇒ an = n² + n + 1
So, the nth term is n² + n + 1

MCQ On Principle Of Mathematical Induction Question 9.
n(n + 1)(n + 5) is a multiple of ____ for all n ∈ N
(a) 2
(b) 3
(c) 5
(d) 7

Hint:
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.

Mathematical Induction MCQ Questions Question 10.
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

Hint:
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
= (n² + 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

Principle Of Mathematical Induction MCQs Question 11.
For any natural number n, 7n – 2n is divisible by
(a) 3
(b) 4
(c) 5
(d) 7

Hint:
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

Principle Of Mathematical Induction MCQ Question 12.
(n² + n) is ____ for all n ∈ N.
(a) Even
(b) odd
(c) Either even or odd
(d) None of these

Hint:
Let P(n): (n² + n) is even.
For n = 1, the given expression becomes (1² + 1) = 2, which is even.
So, the given statement is true for n = 1, i.e., P(1)is true.
Let P(k) be true. Then,
P(k): (k² + k) is even
⇒ (k² + k) = 2m for some natural number m. ….. (i)
Now, (k + 1)² + (k + 1) = k² + 3k + 2
= (k² + k) + 2(k + 1)
= 2m + 2(k + 1) [using (i)]
= 2[m + (k + 1)], which is clearly even.
Therefore, P(k + 1): (k + 1)² + (k + 1) is even
⇒ 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.

Principle Of Mathematical Induction Class 11 Extra Questions Question 13.
For all n ∈ N, 3×52n+1 + 23n+1 is divisible by
(a) 19
(b) 17
(c) 23
(d) 25

Hint:
Given, 3 × 52n+1 + 23n+1
Let n = 1,
3 × 52×1+1 + 23×1+1 = 3 × 52+1 + 23+1 = 3 × 5³ + 24 = 3 × 125 + 16 = 375 + 16 = 391
Which is divisible by 17
Let n = 2,
3 × 52×2+1 + 23×2+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

Maths MCQs For Class 11 With Answers Pdf Question 14.
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

Hint:
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
= (n² + 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

Question 15.
{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: (a) 1/(n + 1) for all n ∈ N.
Hint:
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 16.
(1 + x)n ≥ ____ for all n ∈ N,where x > -1
(a) 1 + nx
(b) 1 – nx
(c) 1 + nx/2
(d) 1 – nx/2

Hint:
Let P(n): (1 + x) )n ≥ (1 + nx).
For n = 1, we have LHS = (1 + x))1 = (1 + x), and
RHS = (1 + 1 ∙ x) = (1 + x).
Therefore LHS ≥ RHS is true.
Thus, P(1) is true.
Let P(k) is true. Then,
P(k): (1 + x)1 ≥ (1 + kx). …….. (i)
Now,(1 + x)k+1 = (1 + x)k (1 + x)
≥ (1 + kx)(1 + x) [using (i)]
=1 + (k + 1)x + kx²
≥ 1 + (k + 1)x + x [Since kx² ≥ 0]
Therefore P(k + 1) : (1 + x)k + 1 ≥ 1 + (k + 1)x
⇒ 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 17.
102n-1 + 1 is divisible by ____ for all N ∈ N
(a) 9
(b) 10
(c) 11
(d) 13

Hint:
Let P (n): (102n-1 + 1) is divisible by 11.
For n=1, the given expression becomes {10(2×1-1) + 1} = 11, which is divisible by 11.
So, the given statement is true for n = 1, i.e., P (1) is true.
Let P(k) be true. Then,
P(k): (102k-1 + 1) is divisible by 11
⇒ (102k-1 + 1) = 11 m for some natural number m.
Now, {102(k-1)-1 – 1 + 1} = (102k+1 + 1) = {10² ∙ 10(2k+1)+ 1}
= 100 × {102k-1 + 1 } – 99
= (100 × 11 m) – 99
= 11 × (100 m – 9), which is divisible by 11
⇒ P (k + 1) : {102(k-1) – 1 + 1} is divisible by 11
⇒ 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 all n∈N, 72n − 48n−1 is divisible by :
(a) 25
(b) 2304
(c) 1234
(d) 26

Hint:
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

Question 19.
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: (d) n(n + 1)(2n + 1)/6
Hint:
Given, series is 1² + 2² + 3² + ………..n²
Sum = n(n + 1)(2n + 1)/6

Question 20.
{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)}

Hint:
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.

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