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Principle of Mathematical Induction Class 11 MCQs Questions with Answers
Solving the MCQ Questions of Principle of Mathematical Induction Class 11 with answers can help you understand the concepts better.
The Principle of Mathematical Induction
Suppose there is a given statement P(n) involving the natural number n such that
(i) The statement is true for n = 1, i.e., P(1) is true, and
(ii) If the statement is true for n = k (where k is some positive integer), then the statement is also true for n = k + 1, i.e., truth of P(k) implies the truth of P (k + 1).
Then, P(n) is true for all natural numbers n.
Property (i) is simply a statement of fact. There may be situations when a statement is true for all n ≥ 4. In this case, step 1 will start from n = 4 and we shall verify the result for n = 4, i.e., P(4).
Property (ii) is a conditional property. It does not assert that the given statement is true for n = k, but only that if it is true for n = k, then it is also true for n = k +1. So, to prove that the property holds, only prove that conditional proposition:
If the statement is true for n = k, then it is also true for n = k + 1.
This is sometimes referred to as the inductive step. The assumption that the given statement is true for n = k in this inductive step is called the inductive hypothesis.
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
Question 11.
(1² + 2² + …… + n²) _____ for all values of n ∈ N
(a) = n³/3
(b) < n³/3
(c) > n³/3
(d) None of these
Answer
Answer: (c) > n³/3
Let P(n): (1² + 2² + ….. + n²) > n³/3.
When = 1, LHS = 1² = 1 and RHS = 1³/3 = 1/3.
Since 1 > 1/3, it follows that P(1) is true.
Let P(k) be true. Then,
P(k): (1² + 2² + ….. + k² ) > k³/3 …. (i)
Now,
1² + 2² + ….. + k²
+ (k + 1)²
= {1² + 2² + ….. + k² + (k + 1)²
> k³/3 + (k + 1)³ [using (i)]
= 1/3 ∙ (k³ + 3 + (k + 1)²) = 1/3 ∙ {k² + 3k² + 6k + 3}
= 1/3[k³ + 1 + 3k(k + 1) + (3k + 2)]
= 1/3 ∙ [(k + 1)³ + (3k + 2)]
> 1/3(k + 1)³
P(k + 1):
1² + 2² + ….. + k² + (k + 1)²
> 1/3 ∙ (k + 1)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 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
Answer: (c) n/{3(2n + 3)}
Let the given statement be P(n). Then,
P(n): {1/(3 ∙ 5) + 1/(5 ∙ 7) + 1/(7 ∙ 9) + ……. + 1/{(2n + 1)(2n + 3)} = n/{3(2n + 3).
Putting n = 1 in the given statement, we get
and LHS = 1/(3 ∙ 5) = 1/15 and RHS = 1/{3(2 × 1 + 3)} = 1/15.
LHS = RHS
Thus, P(1) is true.
Let P(k) be true. Then,
P(k): {1/(3 ∙ 5) + 1/(5 ∙ 7) + 1/(7 ∙ 9) + …….. + 1/{(2k + 1)(2k + 3)} = k/{3(2k + 3)} ….. (i)
Now, 1/(3 ∙ 5) + 1/(5 ∙ 7) + ..…… + 1/[(2k + 1)(2k + 3)] + 1/[{2(k + 1) + 1}2(k + 1) + 3
= {1/(3 ∙ 5) + 1/(5 ∙ 7) + ……. + [1/(2k + 1)(2k + 3)]} + 1/{(2k + 3)(2k + 5)}
= k/[3(2k + 3)] + 1/[2k + 3)(2k + 5)] [using (i)]
= {k(2k + 5) + 3}/{3(2k + 3)(2k + 5)}
= (2k² + 5k + 3)/[3(2k + 3)(2k + 5)]
= {(k + 1)(2k + 3)}/{3(2k + 3)(2k + 5)}
= (k + 1)/{3(2k + 5)}
= (k + 1)/[3{2(k + 1) + 3}]
= P(k + 1): 1/(3 ∙ 5) + 1/(5 ∙ 7) + …….. + 1/[2k + 1)(2k + 3)] + 1/[{2(k + 1) + 1}{2(k + 1) + 3}]
= (k + 1)/{3{2(k + 1) + 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 n ∈ N.
Question 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
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 14.
(2 ∙ 7N + 3 ∙ 5N – 5) is divisible by ……….. for all N ∈ N
(a) 6
(b) 12
(c) 18
(d) 24
Answer
Answer: (d) 24
Let P(n): (2 ∙ 7ⁿ + 3 ∙ 5ⁿ – 5) is divisible by 24.
For n = 1, the given expression becomes (2 ∙ 71 + 3 ∙ 51 – 5) = 24, which is clearly divisible by 24.
So, the given statement is true for n = 1, i.e., P(1) is true.
Let P(k) be true. Then,
P(k): (2 ∙ 7ⁿ + 3 ∙ 5ⁿ – 5) is divisible by 24.
⇒ (2 ∙ 7ⁿ + 3 ∙ 5ⁿ – 5) = 24m, for m = N
Now, (2 ∙ 7ⁿ + 3 ∙ 5ⁿ – 5)
= (2 ∙ 7k ∙ 7 + 3 ∙ 5k ∙ 5 – 5)
= 7(2 ∙ 7k + 3 ∙ 5k – 5) = 6 ∙ 5k + 30
= (7 × 24m) – 6(5k – 5)
= (24 × 7m) – 6 × 4p, where (5k – 5) = 5(5k-1 – 1) = 4p
[Since (5k-1 – 1) is divisible by (5 – 1)]
= 24 × (7m – p)
= 24r, where r = (7m – p) ∈ N
⇒ P (k + 1): (2 ∙ 7k + 13 ∙ 5k + 1 – 5) is divisible by 24.
⇒ 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 15.
For all n∈N, 52n − 1 is divisible by
(a) 26
(b) 24
(c) 11
(d) 25
Answer
Answer: (b) 24
Given number = 52n − 1
Let n = 1, 2, 3, 4, ……..
52n − 1 = 5² − 1 = 25 – 1 = 24
52n − 1 = 54 – 1 = 625 – 1 = 624 = 24 × 26
52n − 1 = 56 – 1 = 15625 – 1 = 15624 = 651 × 24
Since, all these numbers are divisible by 24 for n = 1, 2, 3, …..
So, the given number is divisible by 24
Question 16.
1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ….. + n(n + 1) =
(a) n(n + 1)(n + 2)
(b) {n(n + 1)(n + 2)}/2
(c) {n(n + 1)(n + 2)}/3
(d) {n(n + 1)(n + 2)}/4
Answer
Answer: (c) {n(n + 1)(n + 2)}/3
Let the given statement be P(n). Then,
P(n): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ….. + n(n + 1) = (1/3){n(n + 1) (n + 2)}
Thus, the given statement is true for n = 1, i.e., P(1) is true.
Let P(k) be true. Then,
P(k): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ….. + k(k + 1) = (1/3){k(k + 1) (k + 2)}.
Now, 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 +…+ k(k + 1) + (k + 1) (k + 2)
= (1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ……. + k(k + 1)) + (k + 1) (k + 2)
= (1/3) k(k + 1) (k + 2) + (k + 1)(k + 2) [using (i)]
= (1/3) [k(k + 1) (k + 2) + 3(k + 1)(k + 2)
= (1/3){(k + 1) (k + 2)(k + 3)}
⇒ P(k + 1): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 +……+ (k + 1) (k + 2)
= (1/3){k + 1 )(k + 2) (k +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 values of ∈ N.
Question 17.
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
Answer: (a) {n(n + 3)}/{4(n + 1)(n + 2)}
Let P (n): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……. + 1/{n(n + 1)(n + 2)} = {n(n + 3)}/{4(n + 1)(n + 2)}
Putting n = 1 in the given statement, we get
LHS = 1/(1 ∙ 2 ∙ 3) = 1/6 and RHS = {1 × (1 + 3)}/[4 × (1 + 1)(1 + 2)] = ( 1 × 4)/(4 × 2 × 3) = 1/6.
Therefore LHS = RHS.
Thus, the given statement is true for n = 1, i.e., P(1) is true.
Let P(k) be true. Then,
P(k): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……… + 1/{k(k + 1) (k + 2)} = {k(k + 3)}/{4(k + 1) (k + 2)}. …….(i)
Now, 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………….. + 1/{k(k + 1) (k + 2)} + 1/{(k + 1) (k + 2) (k + 3)}
= [1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………..…. + 1/{ k(k + 1) (k + 2}] + 1/{(k + 1)(k + 2) (k + 3)}
= [{k(k + 3)}/{4(k + 1)(k + 2)} + 1/{(k + 1)(k + 2)(k + 3)}] [using(i)]
= {k(k + 3)² + 4}/{4(k + 1)(k + 2) (k + 3)}
= (k³ + 6k² + 9k + 4)/{4(k + 1) (k + 2) (k + 3)}
= {(k + 1) (k + 1) (k + 4)}/{4 (k + 1) (k + 2) (k + 3)}
= {(k + 1) (k + 4)}/{4(k + 2) (k + 3)
⇒ P(k + 1): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……….….. + 1/{(k + 1) (k + 2) (k + 3)}
= {(k + 1) (k + 2)}/{4(k + 2) (k + 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 18.
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 19.
The sum of n terms of the series 1² + 3² + 5² +……… is
(a) n(4n² – 1)/3
(b) n²(2n² + 1)/6
(c) none of these.
(d) n²(n² + 1)/3
Answer
Answer: (a) n(4n² – 1)/3
Let S = 1² + 3² + 5² +………(2n – 1)²
⇒ S = {1² + 2² + 3² + 4² ………(2n – 1)² + (2n)²} – {2² + 4² + 6² +………+ (2n)²}
⇒ S = {2n × (2n + 1) × (4n + 1)}/6 – {4n × (n + 1) × (2n + 1)}/6
⇒ S = n(4n² – 1)/3
Question 20.
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 + 5n³ + 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
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