Prove using mathematical induction that $2^{3n}-1$ is divisible by $7$ So, i wanna prove $2^{3n}-1$ is divisible by $7$, so i made this:
$2^{3n}-1 = 7\cdot k$ -> for some $k$ value
$2^{3n+1} = 1+2\cdot1 - 2\cdot1 $
$2^{3n+1} - 1-2\cdot1 + 2\cdot1 $
$2^{3n}\cdot2 - 1-2\cdot1 + 2\cdot1$
$2(2^{3n}-1) -1 +2$
$2\cdot7k+1$ -> made this using the hypothesis.
so, i dont know if its right, or if its wrong, i dont know how to keep going from this, or if its the end.
Thanks.

If $\displaystyle f(m)=2^{3m}-1$
$\displaystyle f(m+1)=2^{3(m+1)}-1$ and not $2^{3m+1}-1$
So, $2^{3(m+1)}-1=2^3\cdot2^{3m}-1=2^3(2^{3m}-1)-1+2^3$
