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.

 A: Hint: $6\equiv 1\mod 5$.
Prove $6^n\equiv 1 \mod 5$
The term in $5n$ is a distraction.
A: 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 $5$.
A: Assume $6^n+5n+4$ is multiple of $5$.
Let us consider $$6^{n+1}-5(n+1)+4=(5+1)6^n-5n-5+4=5(6^n-1)+(6^n-5n+4)$$
The first summand is multiple of $5$  and the second is multiple of $5$ by inductive assumtion.
A: By the binomial theorem, $6^n=(1+5)^n=1+5n+5^2a$, and so $6^n-5n+4=5+5^2a$.
A: prove that $6^n-5n+4$ is divisible by 5.
1) Prove true for n=1
$6^1-5(1)+4= 5$, which is divisible by 5. Hence, the preposition is true for n=1.
2) Assume true for n=k, given true for n=1.
$6^k-5(k)+4= 5a$, where a is some integer.
3) Prove true for n=k+1, assuming true for n=k.
$6^k+1 -5(k+1)+4 =5b$, where b is some integer.
$=6^k.6^1 -5k-5+4$, ($x^a+b$ can be written as $X^a.x^b$ , where $6^1=6$.)
$=6.6^k-5k+4-5$ (some simple rearranging)
=6.5a -5  (from assumption that $6^k-5k+4=5a$)
=5(a+6-1) -factorising by taking 5 out common.
= 5(a+5) ---> if multiplied by 5, it must be then divisible by 5.
Therefore the preposition is true for n=k+1, assuming true for n=k and proven true for n=1. This completes the proof.
