Prove by the Principle of Mathematical Induction: $4^n-1$ is divisible by 3, for each natural number n.

Assume given statement

Let $P\left(n\right):4^n-1$ is divisible by 3 for each natural number n.

Check that statement is true for$n=1$

$P\left(1\right)=4^1-1=3$

$\therefore P\left(1\right)$ is divisible by 3.

So, $P\left(1\right)$is true.

Assume P(k) to be true and then prove $P(k+1)$ is true.

Assume that P(n) is true for some$n=kϵN.$

$\therefore P\left(k\right):4^k-1$ is divisible by 3 is true.

i.e.,$4^k-1=3q,$ where $qϵN$

$\Rightarrow 4^k=3q+1\dots \left(1\right)$

Now we have to prove for $P(k+1)$

$P(k+1):4^{k+1}-1={4.4}^k-1$

$=4(3q+1)-1$ (From (1))

$=12q+4-1=3(4q+1),$

where $qϵN\Rightarrow P(k+1)$ is divisible by 3

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

Hence, By Principle of mathematical Induction, P(n) is true for all natural numbers n.