Inductive proof of $\,9 \mathrel| 4^n+6n-1\,$ for all $\,n\in\Bbb N$ Prove that for all $n\in\mathbb N$, $9 \mathrel| (4^n+6n-1)$.
I have shown the base case if $n=0$ (which is a natural number for my course).
I'm now trying to prove that $9k=4^n+6n-1$.
I substituted $n+1$ for $n$, and have $4^{(n+1)}+6(n+1)-1$, which equals $4*4^n+6n+5$.
I'm stuck. Is that not the right way to go? I don't know how to get from the “$+5$” to “$-1$” so I can substitute $9k$ to the right side of the equation. Any help would be greatly appreciated.

It's really simple this is how I'd approach it:
$4^{n+1}+6(n+1)-1=4(4^n+6n-1)-18n+9=4(4^n+6n-1)-9(2n-1)$
and since you've already proved that $4^n+6n-1=9k$ then you will have: $4(4^n+6n-1)-9(2n-1)=9k*4-9(2n-1)$
$=9(4k-(2n-1))$ and since $k$ and $2n-1$ are both integers then you can let $k'=4k-(2n-1)$ .
Now you'll have that $4^{n+1}+6(n+1)-1=9kk'$ which is a multiple of 9. So you can conclude for all natural numbers $n$ that $4^n+6n-1$ is divisible by $9$. 
