Prove by induction that $3^{2n+3}+40n-27$ is divisible by 64 for all n in natural numbers I cannot complete the third step of induction for this one. The assumption is $3^{2n+3}+40n-27=64k$, and when substituting for $n+1$ I obtain $3^{2n+5}+40n+13=64k$. I've tried factoring the expression, dividing, etc. but I cannot advance from here.

Let $A_n =  3^{2n+3}+40n-27$, then $A_n = 11A_{n-1} - 19A_{n-2} + 9A_{n-3}$.
From this it's clear that if 64 divides $A_n$ three consecutive values of $n$ then it holds for the next. So by induction it's enough to check it for $n = -1,0,1$, which is easy enough to do by hand. 
