**To Prove:**

4^{n} + 15n – 1 is divisible by 9.

Let us prove this question by principle of mathematical induction (PMI) for all natural numbers

Let P(n):4^{n} + 15n – 1 is divisible by 9.

For n = 1 P(n) is true since 4^{n} + 15n – 1 = 4^{1} + 15 x 1 – 1 = 18

Which is divisible of 9

Assume P(k) is true for some positive integer k , ie,

= 4^{k} + 15k – 1 is divisible by 9

∴ 4^{k} + 15k – 1 = m x 9, where m ∈ N …(1)

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

Consider,

= 4^{k+1} + 15(k + 1) – 1

= 4^{k} x 4 + 15k + 15 - 1

= 4^{k} x 4 + 15k + 14 + (60k + 4) - (60k + 4)

[Adding and subtracting 60k + 4]

= 9 x r , where r = [(4m) - (5k - 2)] is a natural number

Therefore 4^{k} + 15^{k} - 1 is a divisible of 9

Therefore, P (k + 1) is true whenever P(k) is true

By the principle of mathematical induction, P(n) is true for all natural numbers, ie, N.

**Hence proved.**