NCERT Solutions for class 11 Maths | Chapter 4 - Principles of Mathematical Induction

(1) 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
Answer: n(n+1)(n+2)/6
(2) (1 + x)n ≥ ____ for all n ∈ N,where x > -1
[A] 1 + nx
[B] 1 – nx
[C] 1 + nx/2
[D] 1 – nx/2
Answer: 1 + nx

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(3) For any natural number n, 7n – 2n is divisible by
[A] 3
[B] 4
[C] 5
[D] 7
Answer: 5
(4) 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: n² + n + 1
(5) (n² + n) is ____ for all n ∈ N.
[A] Even
[B] odd
[C] Either even or odd
[D] None of these
Answer: Even
(6) For any natural number n, 7n – 2n is divisible by
[A] 3
[B] 4
[C] 5
[D] 7
Answer: 5
(7) 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: {n(n + 1)/2}²
(8) n(n + 1)(n + 5) is a multiple of ____ for all n ∈ N
[A] 2
[B] 3
[C] 5
[D] 7
Answer: 3
(9) 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: n(n + 1)(2n + 1)/6
(10) For all n ∈ N, 3×52n+1 + 23n+1 is divisible by
[A] 19
[B] 17
[C] 23
[D] 25
Answer: 17
(11) 102n-1 + 1 is divisible by ____ for all N ∈ N
[A] 9
[B] 10
[C] 11
[D] 13
Answer: 11
(12) {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)}
Answer: n/{3(2n + 3)}
(13) 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: a + b
(14) 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: n(n + 1)(2n + 1)/6
(15) For all n∈N, 72n − 48n−1 is divisible by :
[A] 25
[B] 2304
[C] 1234
[D] 26
Answer: 2304
(16) {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: 1/(n + 1) for all n ∈ N.
(17) 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)
Answer: n/(n + 1)
(18) {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: 1/(n + 1) for all n ∈ N.
(19) 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: {n(n + 3)}/{4(n + 1)(n + 2)}
(20) 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
Answer: n(n+1)(n+2)/6

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