<s>[INST] Proof by Induction Divisibility. $6^n-5n+4$ is divisible by 5 for all positive integers $n$.
$n >=1$

Prove By Induction
My attempt is as follows:

$n=1$
  $6^1-5(1) +4$
  $=5$, Therefore 5 is divisible by 5 so $n=1$ is true
  
  Assume its true for $n=k$
  consider $n=k+1$
  $6^k-5k+4=5.x$
I am stuck here would appreciate some assistance.

 [/INST] Assume $6^n+5n+4$ is a multiple of $5$, so
$$
6^n+5n+4=5x
$$
for some integer $x$.
Now you want to do the inductive step; since $6^{n+1}$ seems to be the toughest term, we isolate $6^n$ from the previous identity:
$$
6^n=5x-5n-4
$$
and recall that $6^{n+1}=6\cdot6^n$; then
\begin{align}
6^{n+1}+5(n+1)+4
&=6\cdot6^n+5(n+1)+4\\
&=6(5x-5n-4)+5(n+1)+4\\
&=5(6x-6n+n+1)-24+4\\
&=5(6x-5n+1-4)\\
&=5(6x-5n-3)
\end{align}
is a multiple of $5$.
</s>