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.
 A: (Without induction)
There is a very useful identity $a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \dots + a b^{n-2} + b^{n-1})$.
If you take $a = 2^{3} = 8$ and $b=1$, the result becomes obvious.
A: Assume $2^{3k}-1=7m$ 
Consider $2^{3(k+1)}-1=2^{3k}.2^3-1=(7m+1)8-1=7(8m+1)=7n$
A: If $\displaystyle f(m)=2^{3m}-1$
$\displaystyle f(m+1)=2^{3(m+1)}-1$ and not $2^{3m+1}-1$
So, $2^{3(m+1)}-1=2^3\cdot2^{3m}-1=2^3(2^{3m}-1)-1+2^3$
A: $f(n) = 2^{3n}-1 $
$f(0) = 0$ and $7|0$
Suppose that $7|f(n)$, let's say $f(n) = 7k$, $\Rightarrow f(n+1) = 2^{3n+3}-1 = 8\cdot2^{3n}-1 = 8\cdot2^{3n}-1 + 8 - 8 = 8(2^{3n}-1) - 7 = 8\cdot(7k) - 7 = 7\cdot(8k-1)$
A: $2^{3n}-1=8^n-1=(7+1)^n-1$. Now note that all terms in the expansion of $(7+1)^n$ are multiples of $7$, except the last, which cancels with $-1$. Hence  $2^{3n}-1$ is a multiple of $7$.
A: P(n) = (2^3n)-1
p(1) = (2^3x1)-1
        = 8-1 = 7 is divisible by 7
Let n =k
 then p(k) =( 2^3k)-1 = 7x [assume it is divisible by 7]
 therefore 2^3k = 7x +1 

Let n =k+1
then p(k+1) = (2^3(k+1))-1
            = (2^3k+3)-1
         = (2^3k x 2^3)-1

         = ((7x +1)2^3 ) -1

         = ((7x +1)8 ) -1

          = (56x +8) -1

         = 56x +8 -1

          =56x +7

          = 7(8x +1)

     therefore it is divisible by 7

