Question

Answer the following:

Prove by method of induction 15^2n–1 + 1 is divisible by 16, for all n ∈ N.

Answer 1

15^2n–1 + 1 is divisible by 16, if and only if (15^2n–1 + 1) is a multiple of 16

Let P(n) ≡ 15^2n–1 + 1 = 16m, where m ∈ N.

Step I:

Put n = 1

∴ 15^2n–1 + 1 = `15^(2(1)–1)` + 1 = 15 + 1 = 16 = 16(1) which is a multiple of 16

∴ P(n) is true for n = 1

Step II:

Let us consider that P(n) is true for n = k

i.e., 15^2k–1 + 1 is a multiple of 16

∴ 15^2k–1 + 1 = 16a, where a ∈ N

∴ 15^2k–1 = 16a – 1

∴ `15^(2"k")/15` = 16a – 1

∴ 15^2k = 15.(16a – 1)   ...(i)

Step III:

We have to prove that P(k + 1) is true

i.e., to prove that

`15^(2("k"+1)-1)` + 1 is a multiple of 16.

i.e., `15^(2("k"+1)-1)` + 1 = 16b, where b ∈ N

∴ `15^(2("k"+1)-1)` + 1 

= 15^2k+1 + 1

= 15^2k .15 + 1

= 15(16a – 1)  × 15 + 1   ...[From (i)]

= 225 × 16a – 225 + 1

= 225 × 16a – 224

= 16(225a – 14)

= 16 b, where b = (225a – 14) ∈ N.

∴ P(n) is true for n = k + 1

Step IV:

From all steps above by the principle of mathematical induction, P(n) is true for all n ∈ N.

∴ 15^2n–1 + 1 is divisible by 16, for all n ∈  N.