$$P(n):2^{3n}-1$$

If It's divisible by $$7$$ then,

Base case:

$$n=1$$ then,

$$P(n):2^{3n}-1=2^{3\times1}-1=7$$

If $$n=k$$ is true for some natural numbers, then It should also be true for $$(k+1)$$,

For $$n=k$$,

$$P(k):2^{3k}-1=7d$$

$$P(k):2^{3k}=7d+1$$ .....(1)

Where $$d=1,2,3,4.......so\ on$$

Now we check for $$n=k+1$$,

$$P(k+1):2^{3(k+1)}-1=7d$$

$$P(k+1):2^{3k+3}=7d+1$$

(Since $$a^{xy}=a^x a^y$$)

$$P(k+1):2^{3k}2^3=7d+1$$

Substituting value from equation (1),

$$P(k+1):(7d+1)\times8=7d+1$$

$$ 56d+8=7d+1$$

$$ 7\times(8d+1)=7d$$

So, it is true for both $$k\ and\ (k+1)$$.

Hence, from the principle of mathematical induction,

$$2^{3n}-1$$ is divisible by $$7$$ for all natural numbers.