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
 A: 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:
$=63P+6+21\cdot7^k$
$=63P+6+21(3x+1)$
$=63P+63x+27$
$=9(7P+7x+3)$
A: Another, less elegant way of proving it is using modular arithmetic. In essence, we show that if
$$7^n(3n+1)-1\equiv0 \mod9$$
And, for the inductive step,
$$7^{n+1}(3(n+1)+1)-1=7^{n+1}(3n+4)-1=7(7^n(3n+1)-1+1+3\cdot7^n)-1$$
Using congruence mod $9$:
$$6+3\cdot7^n\equiv0\mod9$$
For all $n\in\mathbb{Z}_9$, which implies what we wanted to show.
A: 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.
A: It can also  be done WITHOUT INDUCTION Il.
$$(3n+1)7^n-1$$
$$=(3n+1)(6+1)^n-1$$
Where in expension of $(6+1)^n$ , except last two terms rest will be multiple of $6^2$
$$=(3n+1)(36m+6n+1)-1 ; where \,\,m\in I $$
$$=108mn+6n+3n +1-1=108mn+9n , where \,\,m,n \in I $$
$$= 9K , where \,\, K \in I $$
