Mathmatical Induction Prove, by mathmatical induction, that $5^{2n}$ + $2^{2n-2}$$3^{n-1}$ is divisible by $13$.
I first plugged in n as 1 and showed that the expression is divisible by 13 for n=1. 
Then I assumed that the expression was divisible by 13 for n=k and plugged in k. The simplified expression in terms of k was $25^k$ + $\frac{4^n3^n}{12}$. I then plugged in k+1 and got an expression for that. I am unable to show that the sum or difference of the k and k+1 expressions is divisible by 13. How do I proceed?

HINT:
For $n=m+1,$
$$f(m+1)=25^{m+1}+4^m3^m=25^{m+1}+12^m$$
$$f(m+2)-25f(m+1)=25^{m+2}+12^{m+1}-25(25^{m+1}+12^m)=-12^m(25-12)$$
$$\implies f(m+2)-25f(m+1)\equiv0\pmod{13}$$
Alternatively, $$f(m+2)-12f(m+1)=25^{m+1}(25-12)\equiv0\pmod{13}$$
By either method, $f(m+2)$ will be divisible by $13$ if $13|f(m+1)$
Now establish the base case i.e., for $n=1$
