This set of Class 11 Maths Chapter 4 Multiple Choice Questions & Answers (MCQs) focuses on “Principle of Mathematical Induction”. These MCQs are created based on the latest CBSE syllabus and the NCERT curriculum, offering valuable assistance for exam preparation.

1. For principle of mathematical induction to be true, what type of number should ‘n’ be?

a) Whole number

b) Natural number

c) Rational number

d) Any form of number

View Answer

Explanation: According to the Principle of Mathematical induction, X(n) can be true if X(1) is true and if X(k) is true. When X(k) is true, it implies that X(k + 1) is also true. Here n can be equal to 0, 1, 2, 3 and so on.

2. 7^{2n} + 2^{2n – 2}. 3^{n – 1} is divisible by 50 by principle of mathematical induction.

a) True

b) False

View Answer

Explanation: P(n) = 7

^{2n}+ 2

^{2n – 2}. 3

^{n – 1}

P(1) = 7

^{2}+ 2

^{0}. 3

^{0}

P(1) = 50

We now assume that P(k) is divisible by 50.

Therefore, P(k) = 7

^{2k}+ 2

^{2k – 2}. 3

^{k – 1}

To prove P(k + 1) = 7

^{2(k – 1)}+ 2

^{2(k + 1) – 2}. 3

^{(k + 1) – 1}is divisible by 50

P(k + 1) = 7

^{2k}. 7

^{2}+ 2

^{2k}. 3

^{k}

P(k + 1) = 7

^{2}( 7

^{2k}+ 2

^{2n – 2}. 3

^{k – 1}– 2

^{2k – 2}. 3

^{k – 1}) + 2

^{2k}. 3

^{k}

P(k + 1) = 7

^{2}x 50c – 7

^{2}. 2

^{2k – 2}. 3

^{k – 1}+ 2

^{2k}. 3

^{k}

Since P(k) = 7

^{2k}+ 2

^{2n – 2}. 3

^{k – 1}– 2

^{2k – 2}. 3

^{k – 1}is divisible by 50, it can be written as 50c.

P(k + 1) = 49 x 50C – 49 . 2

^{2k – 2}. 3

^{k – 1}+ 3 x 4 x 2

^{2k – 2}. 3

^{k – 1}

P(k + 1) = 49 x 50C – 2

^{2k – 2}. 3

^{k – 1}x 37

While the first term is divisible by 50, the second term is not.

Therefore, by principle of mathematical induction, 7

^{2n}+ 2

^{2n – 2}. 3

^{n – 1}is not divisible by 50.

3. By principle of mathematical induction, 2^{4n-1} is divisible by which of the following?

a) 8

b) 3

c) 5

d) 7

View Answer

Explanation:

P(n) = 2

^{4n – 1}

P(1) = 2

^{3}= 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) = 2

^{4}. 2

^{4k – 1}

P(k + 1) = 2

^{4}. 8c

Clearly, P(k + 1) is divisible by 2, 4, 8 and 16.

4. If 10^{3n} + 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

View Answer

Explanation: P(n) = 10

^{3n}+ 2

^{4k + 1}. 9 + k

P(1) = 10

^{3}+ 2

^{5}. 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(n^{2} – 1). Which of the following does not divide P(k+1)?

a) k

b) k + 2

c) k + 3

d) k + 1

View Answer

Explanation: P(n) = n(n

^{2}– 1)

P(k + 1) = (k + 1) ((k + 1)

^{2}– 1)

P(k + 1) = (k + 1) (k

^{2}+ 1 + 2k – 1)

P(k + 1) = (k + 1) (k

^{2}+ 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!

View Answer

Explanation: 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) = k^{2} (k + 3) (k^{2} – 1) is true, then what is P(k + 1)?

a) (k + 1)^{2} (k + 3) (k^{2} – 1)

b) (k + 1)^{2} (k + 4) (k^{2} – 1)

c) (k + 1)^{2} (k + 4) k (k + 2)

d) (k + 1) (k + 4) k (k +2)

View Answer

Explanation: 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) (k

^{2}+ 1 + 2k – 1)

P(k + 1) = (k + 1)

^{2}(k + 4) (k

^{2}+ 2k)

P(k + 1) = (k + 1)

^{2}(k + 4) k (k +2)

8. n^{2} + 3n is always divisible by which number, provided n is an integer?

a) 2

b) 3

c) 4

d) 5

View Answer

Explanation: P(n) = n

^{2}+ 3n

P(1) = 1 + 3

P(1) = 4

Let’s assume that P(k) is true and divisible by 4. Therefore, P(k) = k

^{2}+ 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) = k

^{2}+ 1 + 2k + 3k + 3

P(k+1) = k

^{2}+ 5k + 4

P(k+1) = (k

^{2}+ 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. n^{3} + 5n is divisible by which of the following?

a) 3

b) 5

c) 7

d) 11

View Answer

Explanation: P(n) = n

^{3}+ 5n

P(1) = 1 + 5

P(1) = 6

We assume the P(k) is true and divisible by 6.

P(k) = k

^{3}+ 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) = k

^{3}+ 1 + 3k

^{2}+ 3k + 5k + 5

P(k + 1) = (k

^{3}+ 5k) + 3k

^{2}+ 3k + 6

P(k + 1) = 6c + 3(k

^{2}+ k + 2)

P(k + 1) = (3 x 2c) + 3(k

^{2}+ 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

View Answer

Explanation: 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) = n^{3} (n + 1)?

a) (k + 1)^{4}

b) k^{4} + 5k^{3} + 9k^{2} + 7k + 2

c) k^{4} + 6k^{3} + 9k^{2} + 7k + 2

d) k^{4} + 3k^{3} + 9k^{2} + 6k + 2

View Answer

Explanation: P(n) = n

^{3}(n + 1)

P(k + 1) = (k + 1)

^{3}(k + 1 + 1)

P(k + 1) = (k

^{3}+ 3k

^{2}+ 3k + 1) (k + 2)

P(k + 1) = k

^{4}+ 3k

^{3}+ 3k

^{2}+ k + 2k

^{3}+ 6k

^{2}+ 6k + 2

P(k + 1) = k

^{4}+ 5k

^{3}+ 9k

^{2}+ 7k + 2

To practice all chapters and topics of class 11 Mathematics, __ here is complete set of 1000+ Multiple Choice Questions and Answers__.

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