250+ TOP MCQs on The Principle of Mathematical Induction & Answers | Class 11 Maths

Answer: a
Clarification:
P(n) = 2^{4n – 1}
P(1) = 2³ = 8
Let us assume P(k) is divisible by 8 and can be written as 8c, where c is any integer.
P(k) = 2^{4k – 1} = 8c
P(k + 1) = 2^{4(k + 1) – 1}
P(k + 1) = 2^{4k + 3}
P(k + 1) = 2⁴ . 2^{4k – 1}
P(k + 1) = 2⁴ . 8c
Clearly, P(k + 1) is divisible by 2, 4, 8 and 16.

4. If 10³ⁿ + 2^{4k + 1}. 9 + k, is divisible by 11, then what is the least positive value of k?
a) 7
b) 6
c) 8
d) 10
Answer: d
Clarification: P(n) = 10³ⁿ + 2^{4k + 1}. 9 + k
P(1) = 10³ + 2⁵ . 9 + k
P(1) = 1000 + 288 + k
P(1) = 1288 + k
When 1288 is divided by 11, the remainder is 1.
Therefore, 1287 is divisible by 11.
The next number that is divisible is 1298.
k = 1298 – 1288
k = 10

5. P(n) = n(n² – 1). Which of the following does not divide P(k+1)?
a) k
b) k + 2
c) k + 3
d) k + 1
Answer: c
Clarification: P(n) = n(n² – 1)
P(k + 1) = (k + 1) ((k + 1)² – 1)
P(k + 1) = (k + 1) (k² + 1 + 2k – 1)
P(k + 1) = (k + 1) (k² + 2k)
P(k + 1) = (k + 1) k (k + 2)
Therefore, k, (k + 1), (k – 1) divide P(k + 1).

6. What would be the hypothesis of mathematical induction for n(n + 1) < n! (where n ≥ 4) ?
a) It is assumed that at n = k, k(k + 1)! > k!
b) It is assumed that at n = k, k(k + 1)! < k!
c) It is assumed that at n = k, k(k + 1)! > (k + 1)!
d) It is assumed that at n = k, (k + 1)(k + 2)! < k!
Answer: b
Clarification: When we use the principle of mathematical induction, we assume that P(n) is true for P(k) and prove that P(k + 1) is also true. Here P(k) is k(k + 1)! < k!

7. If P(k) = k² (k + 3) (k² – 1) is true, then what is P(k + 1)?
a) (k + 1)² (k + 3) (k² – 1)
b) (k + 1)² (k + 4) (k² – 1)
c) (k + 1)² (k + 4) k (k + 2)
d) (k + 1) (k + 4) k (k +2)
Answer: c
Clarification: In mathematical induction, if P(k) is true, we need to prove that P(k + 1) is also true. Here P(k + 1) is found by substituting (k + 1) in place of k. P(k + 1) = (k + 1)² (k + 1 + 3) ((k + 1)² – 1)
P(k + 1) = (k + 1)² (k + 4) (k² + 1 + 2k – 1)
P(k + 1) = (k + 1)² (k + 4) (k² + 2k)
P(k + 1) = (k + 1)² (k + 4) k (k +2)

8. n² + 3n is always divisible by which number, provided n is an integer?
a) 2
b) 3
c) 4
d) 5
Answer: a
Clarification: P(n) = n² + 3n
P(1) = 1 + 3
P(1) = 4
Let’s assume that P(k) is true and divisible by 4. Therefore, P(k) = k² + 3k can be written as 4c.
We need to check if P(k + 1) is divisible by 4
P(k+1) = (k + 1)² + 3(k + 1)
P(k+1) = k² + 1 + 2k + 3k + 3
P(k+1) = k² + 5k + 4
P(k+1) = (k² + 3k) + 2k + 4
P(k+1) = 4c + 2k + 4
P(k+1) = 4c + 2(k + 2)
Clearly the second part of the equation is not divisible by 4. However P(k) = 4c is divisible by 2 and
P(k + 1) is also divisible by 2. Therefore, 2 divides P(n).

9. n³ + 5n is divisible by which of the following?
a) 3
b) 5
c) 7
d) 11
Answer: a
Clarification: P(n) = n³ + 5n
P(1) = 1 + 5
P(1) = 6
We assume the P(k) is true and divisible by 6.
P(k) = k³ + 5k is divisible by 6 and can be written as 6c or 3 x 2c
We need to prove that P(k + 1) is divisible by 6
P(k + 1) = (k + 1)³ + 5(k + 1)
P(k + 1) = k³ + 1 + 3k² + 3k + 5k + 5
P(k + 1) = (k³ + 5k) + 3k² + 3k + 6
P(k + 1) = 6c + 3(k² + k + 2)
P(k + 1) = (3 x 2c) + 3(k² + k + 2)
Therefore, P(k + 1) is definitely divisible by 3

10. State whether the following series is true or not.
1 + 2 + 3 +…..+ n = (frac{n(n + 1)}{2})
a) True
b) False
Answer: a
Clarification: P(n) = n(n + 1)/2
P(1) = 1
We assume P(k) to be true, therefore, P(k) = (frac{k(k + 1)}{2})
To prove that, P(k + 1) = (frac{(k+1)(k+ 2)}{2})
Proof:
P(k + 1) = 1 + 2 + 3 +….+ k + k + 1
P(k + 1) = (frac{k(k + 1)}{2}) + k+1
P(k + 1) = (frac{k(k+ 1)+2(k+1)}{2})
P(k + 1) = (frac{(k+1)(k+ 2)}{2})
Therefore, P(n) is true by principle of mathematical induction.

11. What will be P(k + 1) for P(n) = n³ (n + 1)?
a) (k + 1)⁴
b) k⁴ + 5k³ + 9k² + 7k + 2
c) k⁴ + 6k³ + 9k² + 7k + 2
d) k⁴ + 3k³ + 9k² + 6k + 2
Answer: b
Clarification: P(n) = n³ (n + 1)
P(k + 1) = (k + 1)³ (k + 1 + 1)
P(k + 1) = (k³ + 3k² + 3k + 1) (k + 2)
P(k + 1) = k⁴ + 3k³ + 3k² + k + 2k³ + 6k² + 6k + 2
P(k + 1) = k⁴ + 5k³ + 9k² + 7k + 2