NCERT solution Class 11 principle of mathematical induction

NCERT Solutions

Class 11 Maths

Principle of Mathematical Induction

NCERT Questions

Ex.4.1 Q.1
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Ex.4.1 Q.24

Ex.4.1 Q.22

Prove the following by using the principle of mathematical induction for all n ∈ N:

3^{2n + 2} – 8n – 9 is divisible by 8.

Let the given statement be P(n), i.e.,

P(n): 3^{2n + 2} – 8n – 9 is divisible by 8.

It can be observed that P(n) is true for n = 1

since 3²^×¹ + 2 – 8 × 1 – 9 = 64, which is divisible by 8.

Let P(k) be true for some positive integer k, i.e.,

3^{2k + 2} – 8k – 9 is divisible by 8.

3^{2k + 2} – 8k – 9 = 8m; where m ∈ N ..............(1)

We shall now prove that P (k + 1) is true whenever P(k) is true.

Consider

3^{2(k + 1) + 2} – 8(k + 1) – 9

= 3^{2k + 2 + 2} – 8k - 8 – 9

= 3^{2k + 2}× 3² – 8k – 9

= 3²(3^{2k + 2} – 8k – 9 + 8k + 9) – 8k - 17

= 3²(3^{2k + 2} – 8k – 9) + 3²(8k + 9) – 8k - 17

= 9 × 8m + 9(8k + 9) – 8k - 17

= 9 × 8m + 72k + 81 – 8k - 17

= 9 × 8m + 64k + 64

= 8(9m + 8k + 8)

= 8r, where r = (9m + 8k + 8) is a natural number.

Therefore, 3^{2(k + 1) + 2} – 8(k + 1) – 9 is divisible by 8.

Thus, P (k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

Principle of Mathematical Induction Class 11 Maths NCERT Chapter 4 NCERT Solutions

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