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

To address your question directly:
You calculations are correct but - as you realized by yourself - the conclusion at the end is still not justified since there are fractions in "$= 9(7P +2/3+21\cdot7^k/9)$".
So, just go one step back and try to squeeze out factor $9$, for example, as follows:
\begin{eqnarray}63P+6+21\cdot7^k
& = & 63P + 3(2+7^{k+1}) \\
& \stackrel{7^{k+1}=(6+1)^{k+1}=6m+1}{=} & 63P + 3(2+1+6m) \\
& = & 63P + 9(1+2m) \\
& = & 9(7P + 1+2m)
\end{eqnarray}
Now, the conclusion is justified.
