Non-homogenous recurrence relation I need to solve the recurrence relation  $A(n)=2A(n-2)+ 2^{n-2}$. I tried writing out equations up to the $A(2)$ and multiplying by powers of two and adding all the equations together  then all the terms cancelled but after that I couldn't find the sum of the powers of two. I forgot to mention that the base is $A(2)=2$ and n is even

It never hurts in such problems to start by gathering some numerical data:
$$\begin{array}{rcc}
n:&2&4&6&8&10&12\\
A(n):&2&8&32&128&512&2048\\
A(n):&2^1&2^3&2^5&2^7&2^9&2^{11}
\end{array}$$
There’s a very obvious pattern here, that leads to the conjecture that $A(n)=2^{\text{what function of }n?}$ if $n$ is even. Complete the conjecture correctly, and you should have little difficulty using mathematical induction to prove that it’s correct.
Alternatively, let $B(n)=A(2n)$; then $$B(n)=2B(n-1)+2^{2(n-1)}=2B(n-1)+4^{n-1}\;,$$
and ‘unwrap’ the recurrence:
$$\begin{align*}
B(n)&=2B(n-1)+4^{n-1}\\
&=2\Big(2B(n-2)+4^{n-2}\Big)+4^{n-1}\\
&=2^2B(n-2)+2\cdot 4^{n-2}+4^{n-1}\\
&=2^2\Big(B(n-3)+4^{n-3}\Big)+2\cdot 4^{n-2}+4^{n-1}\\
&=2^3B(n-3)+2^2\cdot4^{n-3}+2\cdot 4^{n-2}+4^{n-1}\\
&\;\vdots\\
&=2^kB(n-k)+2^{k-1}\cdot 4^{n-k}+2^{k-2}\cdot 4^{n-k+1}+\ldots+2\cdot 4^{n-2}+4^{n-1}\\
&\;\vdots\\
&=2^{n-1}B(1)+\sum_{i=1}^{n-1}2^{i-1}\cdot4^{n-i}\\
&=2^n+\sum_{i=1}^{n-1}2^{i-1}\cdot4^{n-i}\\
&=2^n+\sum_{i=1}^{n-1}2^{i-1}\cdot2^{2n-2i}\\
&=2^n+\sum_{i=1}^{n-1}2^{2n-i-1}\\
&=2^n+\sum_{i=n}^{2n-2}2^i\\
&=2^n+2^n\sum_{i=0}^{n-2}2^i\\
&=2^n\left(1+\sum_{i=0}^{n-2}2^i\right)\\
&=\;?
\end{align*}$$
