Induction exercise check-up Prove by induction on $n$ that $13$ divides $2^{4n+2} + 3^{n+2}$ for all natural $n$.
For base case it is divisble by 13, and 
$2^{4n+6} + 3^{n+3}$ must be divisble too.
$16 * 2^{4n+2}+ 3* 3^{n+2}$
If we denote $2^{4n+2} + 3^{n+2}$ as some $13y$, 
we have $3*13y + 13*2^{4n+2}$
Didn't I make a mistake in assuming that $2^{4n+2} + 3^{n+2}$  is divisble by 13? In induction we have to prove that if something is divisible/whatever by something then n+1 is divisible too which I did

your assumption is necessary in the induction
the steps of induction is :
1. calculate when n = 1 it is true or not
2. assume when n = k is true then check whether n = k + 1 is also true.


*

*for n = 1,


$2^{4n+2}+3^{n+2} = 2^6+3^3 = 7*13$


*assume $2^{4k+2}+3^{k+2}$ is divisible then


$2^{4(k+1)+2}+3^{(k+1)+2}$
$=2^{4k+6}+3^{k+3}$ 
$=16*2^{4k+2}+3*3^{k+2}$ 
$=3*(2^{4k+2}+3^{k+2})+13*2^{4k+2}$ 
$=13x$ for some real x
So for all natural n,   $2^{4k+2}+3^{k+2}$ is divisible by 13. Q.E.D.
