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. 

I would do it as follows.
$$ \sum_{n=0}^{\infty}{\frac{(7n+32)(3^n)}{n(n+2)(4^n)}}=
7\sum_{n=0}^{\infty}{\frac{1}{n+2}{\left(\frac34\right)}^n} +
32\sum_{n=0}^{\infty}{\frac{1}{n(n+2)}{\left(\frac34\right)}^n}$$
Now, we have that
$$\sum_{n=0}^{\infty}{\frac{1}{n+2}{\left(\frac34\right)}^n}=
{\left(\frac34\right)}^{-2}\cdot\sum_{n=0}^{\infty}{\frac{1}{n+2}{\left(\frac34\right)}^{n+2}}\\
={\left(\frac34\right)}^{-2}\cdot\sum_{n=2}^{\infty}{\frac{1}{n}{\left(\frac34\right)}^{n}}$$
Now, for $|r|<1$, we have that
$${(1-r)}^{-1}=\sum_{n\geq0}r^n,$$
so that integrating both sides by $r$ yields
$$-\ln(1-r)=\sum_{n\geq0}\frac1{n+1}r^{n+1}=\sum_{n\geq1}\frac1nr^n.$$
It follows that
$$\sum_{n=2}^{\infty}{\frac{1}{n}{\left(\frac34\right)}^{n}}=-\ln\left(1-\frac34\right)-\frac34=2\ln(2)-\frac34.$$
Finally, we use a partial fraction decomposition on $\frac1{n(n+2)}=\frac12\frac1n-\frac12\frac1{n+2}$, and this splits the final series into two series that can be solved like the one we just did. Do you think you can finish it?
EDIT:
Well, I just noticed that in your opening question there is a division by $0$ in the $n=0$ term, but this shouldn't change your approach to solving the question.
Assuming the summation is carried out for $n\geq1$, you should obtain $\frac{33}2$ in the end.
