**Question:**

Using the principle of mathematical induction, prove each of the following for all n ϵ N:

$\left(4^{n}+15 n-1\right)$ is divisible by 9

**Solution:**

To Prove:

$4^{n}+15 n-1$ is a divisible of 9

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

Let $\mathrm{P}(\mathrm{n}): 4^{n}+15 n-1$ is a divisible of 9

For $n=1 P(n)$ is true since $4^{n}+15 n-1=4^{1}+15 \times 1-1=18$

Which is divisible of 9

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

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

$\therefore 4^{k}+15 k-1=m \times 9$, where $\mathrm{m} \in \mathrm{N} \ldots(1)$

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

Consider,

$=4^{k+1}+15(k+1)-1$

$=4^{k} \times 4+15 k+15-1$

$=4^{k} \times 4+15 k+14+(60 k+4)-(60 k+4)$ [Adding and subtracting $60 k+4]$

$=\left(4^{k+1}+60 k-4\right)+15 k+14-(60 k-4)$

$=4\left(4^{k}+15 k-1\right)+15 k+14-(60 k-4)$

$=4(9 m)-45 k+18[U \operatorname{sing} 1]$

$=4(9 m)-9(5 k-2)$

$=9[(4 m)-(5 k-2)]$

$=9 \times r$, where $r=[(4 m)-(5 k-2)]$ 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.