MCQs For NCERT Class 11 Maths Principle of Mathematical Induction With Answers

Please refer to the MCQ Questions for Class 11 Principle of Mathematical Induction Maths Chapter 4 with Answers. The following Principle of Mathematical Induction Class 11 Mathematics MCQ Questions have been designed based on the latest syllabus and examination pattern for Class 11. Our experts have designed MCQ Questions for Class 11 Principle of Mathematical Induction with Answers for all chapters in your NCERT Class 11 Mathematics book. You can access all MCQs for Class 11 Mathematics

Principle of Mathematical Induction Class 11 MCQ Questions with Answers

See below Principle of Mathematical Induction Class 11 Mathematics MCQ Questions, solve the questions and compare your answers with the solutions provided below.

Question. If 4n/n+1 < (2n)!/(n!)2 , then P(n) is true for
(a) n ≥ 1
(b) n > 0
(c) n < 0
(d) n ≥ 2

D

Question. If n is a natural number, then (n+1/2)≥ n! is true when
(a) n > 1
(b) n ≥ 1
(c) n > 2
(d) n ≥ 2

B

Question. By using principle of mathematical induction for every natural number, (ab)n =
(a) anbn
(b) an b
(c) abn
(d) 1

A

Question. If P(n) = 2 + 4 + 6 + …..+ 2n, n ∈ N, then P(k) = k(k + 1) + 2 ⇒ P(k + 1) = (k + 1)(k + 2) + 2 for all k ∈ N. So we can conclude that P(n) = n(n + 1) + 2 for
(a) all n ∈ N
(b) n > 1
(c) n > 2
(d) nothing can be said

D

Question. If P(n) : 2 + 4 + 6 +… + (2n), n ∈ N, then P(k) = k (k + 1) + 2 implies P (k + 1) = (k + 1) (k + 2) + 2 is true for all k ∈ N. So statement P(n) = n (n + 1) + 2 is true for:
(a) n ≥ 1
(b) n ≥ 2
(c) n ≥ 3
(d) None of these

D

Question. If m, n are any two odd positive integers with n < m, then the largest positive integer which divides all the numbers of the type m2 – n2 is
(a) 4
(b) 6
(c) 8
(d) 9

C

Question. For all n ∈ N, (1+3/1)(1+5/4)(1+7/9)……..(1+(2n+1)/n2) is equal to
(a) (n+1 )2/2
(b) (n+1 )3/3
(c) (n + 1)2
(d) None of these

C

Question. For natural number n, (n!)2 > nn, if
(a) n > 3
(b) n > 4
(c) n ≥ 4
(d) n ≥ 3

D

Question. 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

C

Question. Let S(K) = 1+ 3+ 5…+ (2K − 1) = 3 + k2 , then which of the following is true?
(a) Principle of mathematical induction can be used to prove the formula
(b) S(K) ⇒ S(K + 1)
(c) S(K) ⇒ S(K − 1)
(d) S(1) is correct

B

Question. For all n ∈ N, 1 + 1/1+2 + 1/1+2+3 + ….. + 1/1+2+3+…….+n is equal to
(a) 3n/n + 1
(b) n/n + 1
(c) 2n/n – 1
(d) 2n/n + 1

D

Question. By the principle of induction ∀ n ∈ N, 32n when divided by 8, leaves remainder
(a) 2
(b) 3
(c) 7
(d) 1

D

Question. Let P(n) be 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) None of these

C

Question. For all n ∈ N, the sum of n5/5 + n3/3 + 7n/15 is
(a) a negative integer
(b) a whole number
(c) a real number
(d) a natural number

D

Question. 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

C

Question. If n is a positive integer, then 52n + 2 – 24n – 25 is divisible by
(a) 574
(b) 575
(c) 674
(d) 576

D

Question. The remainder when 54n is divided by 13, is
(a) 1
(b) 8
(c) 9
(d) 10

A

Question. For all n ≥ 1, 12 + 22 + 32 + 42 + ….. + n2 =
(a) n (n+1)/6
(b) n(n+1) (2n–1)
(c) n (n–1) (2n+1)/2
(d) n (n+1) (2n+1)/6

D

Question. For all n ∈ N, 1.3 + 2.32 + 3.33 + ….. + n.3n is equal to

B

ASSERTION – REASON TYPE QUESTIONS

(a) Assertion is correct, Reason is correct; reason is a correct explanation for assertion.
(b) Assertion is correct, Reason is correct; reason is not a correct explanation for assertion
(c) Assertion is correct, Reason is incorrect
(d) Assertion is incorrect, Reason is correct.

Question. Assertion : 11m+ 2 + 122m+ 1 is divisible by 133 for all m ∈ N.
Reason : xn – yn is divisible by x + y, ∀ n ∈ N, x ≠ y.