<s>[INST] 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.
 [/INST] 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

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