Download PDF
Free download in PDF Class 11 Maths Chapter 4 Principles of Mathematical Induction Multiple Choice Questions and Answers for Board, JEE, NEET, AIIMS, JIPMER, IIT-JEE, AIEE and other competitive exams. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. These short solved questions or quizzes are provided by Gkseries./p>
(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
(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
(3)
For any natural number n, 7n – 2n is divisible by
(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
(5)
(n² + n) is ____ for all n ∈ N.
[A]
Even
[B]
odd
[C]
Either even or odd
[D]
None of these
(6)
For any natural number n, 7n – 2n is divisible by
(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
(8)
n(n + 1)(n + 5) is a multiple of ____ for all n ∈ N
(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
(11)
102n-1 + 1 is divisible by ____ for all N ∈ N
[A]
9
[B]
10
[C]
11
[D]
13
(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)}
(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
(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
(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)
(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
Please share this page
Chapters