Practice Questions with Answers - Multiple Choice Questions

MCQ Questions for Class 11 Maths Chapter 4 Principle of Mathematical Induction

MCQ Questions for Class 11 Maths Chapter 4 Principle of Mathematical Induction

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, 3n5 + 5n³ + 7n is divisible by
(a) 5
(b) 15
(c) 10
(d) 3

Answer

Answer: (b) 15
Given number = 3n5 + 5n² + 7n
Let n = 1, 2, 3, 4, ……..
3n5 + 5n³ + 7n = 3 × 1² + 5 × 1³ + 7 × 1 = 3 + 5 + 7 = 15
3n5 + 5n³ + 7n = 3 × 25 + 5 × 2³ + 7 × 2 = 3 × 32 + 5 × 8 + 7 × 2 = 96 + 40 + 14 = 150 = 15 × 10
3n5 + 5n³ + 7n = 3 × 35 + 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, 32n + 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, ……..
32n + 7 = 3² + 7 = 9 + 7 = 16
32n + 7 = 34 + 7 = 81 + 7 = 88
32n + 7 = 36 + 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 = 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


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