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.

HINT:
Let  $\displaystyle f(n)=4^n+6n-1$
So, $\displaystyle f(m+1)-f(m)=4^m(4-1)+6=3(4^m-1)+9$
Using Why $a^n - b^n$ is divisible by $a-b$? $4^m-1$ is divisible by $4-1=3$
So, $\displaystyle f(m+1)-f(m)$ is divisible by $9$
$\displaystyle\implies  f(m+1)$ will be  divisible by $9$ if $\displaystyle f(m)$ is
Now, establish the base case i.e., for $n=1$

As to the origin of the problem,
for $n\in\mathbb N$,  $\displaystyle4^n=(1+3)^n=\sum_{0\le r\le n}\binom nr3^r=1+3n+\sum_{2\le r\le n}\binom nr3^r$ 
Observe that the last part is divisible by $3^2=9$ as Proof that a Combination is an integer
So, $\displaystyle4^n=1+3n+9k$(where $k$ is an integer $\ge0$ ) 
or in congruence form  $\displaystyle4^n\equiv1+3n\pmod9\  \ \ \ (1)$
This can be  derived using induction as follows
If $\displaystyle4^m=1+3\cdot\underbrace m+9\cdot\underbrace  u$ (where $u$ is an integer),
$\displaystyle\implies4^{m+1}=4\cdot4^m=(1+3)(1+3m+9u)=1+3\underbrace{(m+1)}+9\underbrace{(u+3u+m)} $
Now $\displaystyle6n-1$ actually compensates the constant and the coefficient of $n$ in $(1)$
More generally, we can prove $\displaystyle(1+a)^n\equiv1+na\pmod{a^2}$
In that case, the compensator will be $\displaystyle(a^2-a)n-1\pmod{a^2}$
