Infinite series of formula How do I compute this infinite series
$$ \sum_{k=0}^{\infty}{\frac{(7n+32)(3^n)}{n(n+2)(4^n)}}$$
I believe partial fraction decomposition is part of the solving method but am a little stuck because the $3^n$ and the $4^n$ make the decomposition method a bit strange:
${\frac{(7n+32)(3^n)}{n(n+2)(4^n)}} = {\frac{A}{n}} + {\frac{B}{n+2}} + {\frac{C}{4^n}}$
$(7n+32)(3^n) = A(n+2)(4^n) + B(n)(4^n) + C(n)(n+2)$
$7n(3^n) + 32(3^n)  = An (4^n) + 2A(4^n) +  B(n)(4^n) + C(n^2 + 2n)$
$7n(3^n) + 32(3^n)  = (An +2A + Bn)(4^n) + C(n^2 + 2n)$
..Stuck on how to proceed. 

HINT:
$$\frac{(7n+32)(3^n)}{n(n+2)(4^n)}=\frac{(4^2(n+2)-3^2)(3^n)}{n(n+2)(4^n)}$$
$$=\dfrac{3^n}{n4^{n-2}}-\dfrac{3^{n+2}}{(n+2)4^n}$$
$$=16\left(\dfrac{(3/4)^n}n-\dfrac{(3/4)^{n+2}}{n+2}\right)$$
