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.


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.
