Prove by mathematical induction that $(3n+1)7^n -1$ is divisible by $9$ for integral $n>0$ $7^n(3n+1)-1=9m$
$S_k = 7^k(3k+1)-1=9P$
$\Rightarrow 7^k(3k+1) = 9P+1$
$S_{k+1} = 7\cdot7^k(3(k+1)+1)-1$
$= 7\cdot7^k(3k+1+3)-1$
$= 7\cdot7^k(3k+1) +21\cdot7^k -1$
$= 7(9P+1)+21\cdot7^k -1$
$= 63P+7+21\cdot7^k -1$
$= 63P+6+21\cdot7^k$
$= 9(7P +2/3+21\cdot7^k/9)$
therefore it is divisible by $9$
So I believe I have done this right but I've ended up with non-integers in the answer which im pretty sure isn't right.
Where have I gone wrong?
Thanks

You're almost finished. 
You assume that 
$S_k = 7^k(3k+1)-1$ is divisible by 9, and therefore divisible by 3.
$S_k = 7^k(3k)+ 7^k-1$ is divisible by 3.
Therefore $7^k-1$ is divisible by 3. 
let $3x = 7^k-1$. 
$7^k=3x+1$
Then in your final steps:
$=63P+6+21\cdot7^k$
$=63P+6+21(3x+1)$
$=63P+63x+27$
$=9(7P+7x+3)$
