MCQ Questions Chapter 4 Principle of Mathematical Induction Class 11 Mathematics

MCQ Class 11

Please refer to MCQ Questions Chapter 4 Principle of Mathematical Induction Class 11 Mathematics with answers provided below. These multiple-choice questions have been developed based on the latest NCERT book for class 11 Mathematics issued for the current academic year. We have provided MCQ Questions for Class 11 Mathematics for all chapters on our website. Students should learn the objective based questions for Chapter 4 Principle of Mathematical Induction in Class 11 Mathematics provided below to get more marks in exams.

Chapter 4 Principle of Mathematical Induction MCQ Questions

Please refer to the following Chapter 4 Principle of Mathematical Induction MCQ Questions Class 11 Mathematics with solutions for all important topics in the chapter.

MCQ Questions Answers for Chapter 4 Principle of Mathematical Induction Class 11 Mathematics

Question. The statement P(n)
“1 × 1! + 2 × 2! + 3 × 3! + ….. + n × n! = (n + 1)! – 1” is
(a) True for all n > 1
(b) Not true for any n
(c) True for all n ∈ N
(d) None of these

Answer

C

Question. The greatest positive integer, which divides (n + 1) (n + 2) (n + 3) ….. (n + r) for all n ∈ W, is
(a) r
(b) r!
(c) n + r
(d) (r + 1)!

Answer

B

Question. For natural number n, 2n (n – 1)! < nn, if
(a) n < 2
(b) n > 2
(c) n ≥ 2
(d) n > 3

Answer

B

Question. Use principle of mathematical induction to find the value of k, where (102n – 1 + 1) is divisible by k.
(a) 11
(b) 12
(c) 13
(d) 9

Answer

A

Question. For given series:
12 + 2 × 22 + 32 + 2 × 42 + 52 + 2 × 62 + ….., if Sn is the sum of n terms, then
(a) Sn = (n+1 )2/2 , if n is even
(b) Sn = n2(n+1) /2 , if n is odd
(c) Both (a) and (b) are true
(d) Both (a) and (b) are false

Answer

C

Question. By mathematical induction

MCQ Questions Chapter 4 Principle of Mathematical Induction Class 11 Mathematics
Answer

B

Question. If n ∈ N, then 11n + 2 + 122n + 1 is divisible by
(a) 113
(b) 123
(c) 133
(d) None of these

Answer

C

Question. Let P(n) : n2 + n + 1 is an even integer. If P(k) is assumed true then P(k + 1) is true. Therefore P(n) is true.
(a) for n > 1
(b) for all n ∈ N
(c) for n > 2
(d) None of these

Answer

D

Question. If P(n) : “46n + 16n + k is divisible by 64 for n ∈ N” is true, then the least negative integral value of k is.
(a) – 1
(b) 1
(c) 2
(d) – 2

Answer

A

Question. For all n ∈ N, 41n – 14n is a multiple of
(a) 26
(b) 27
(c) 25
(d) None of these

Answer

B

Question. For all n ∈ N, 3.52n + 1 + 23n + 1 is divisible by
(a) 19
(b) 17
(c) 23
(d) 25

Answer

B

Question. Let P(n) : “2n < (1 × 2 × 3 × … × n)”. Then the smallest positive integer for which P(n) is true is
(a) 1
(b) 2
(c) 3
(d) 4

Answer

D

Question. Statement-I : 1 + 2 + 3 + ….. + n < 1/8 (2n + 1)2, n ∈ N.
Statement-II : n(n + 1) (n + 5) is a multiple of 3, n ∈ N.
(a) Only Statement I is true
(b) Only Statement II is true
(c) Both Statements are true
(d) Both Statements are false

Answer

C

Question. For all n ≥ 1, 1/1.2 + 1/2.3 + 1/3.4 + …..+ 1/n (n+1)=
(a) n
(b) n/n+1
(c) (n+1)/n
(d) 4n+3/2n

Answer

B

Question. 10n + 3(4n + 2) + 5 is divisible by (n ∈ N)
(a) 7
(b) 5
(c) 9
(d) 17

Answer

C

Question. When 2301 is divided by 5, the least positive remainder is
(a) 4
(b) 8
(c) 2
(d) 6

Answer

C

Question. 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

Answer

B

Question. For every positive integer n, 7n – 3n is divisible by
(a) 7
(b) 3
(c) 4
(d) 5

Answer

C

Question. P(n) : 2.7n + 3.5n – 5 is divisible by
(a) 24, ∀ n ∈ N
(b) 21, ∀ n ∈ N
(c) 35, ∀ n ∈ N
(d) 50, ∀ n ∈ N

Answer

A

Question. Principle of mathematical induction is used
(a) to prove any statement
(b) to prove results which are true for all real numbers
(c) to prove that statements which are formulated in terms of n, where n is positive integer
(d) in deductive reasoning

Answer

C