How could I find the sum of this infinite series by hand? $$\sum_{n=1}^{\infty}\frac{(7n+32)3^n}{n(n+2)4^n}$$
Thank you!
 A: HINT
Notice, $$\frac{(7n+32)3^n}{n(n+2)4^n}=\left(\frac{3}{4}\right)^n\frac{(7n+32)}{n(n+2)}=\left(\frac{3}{4}\right)^n\left(\frac{16}{n}-\frac{9}{n+2}\right)$$
A: $$\sum_{n\geq 1}\frac{z^{n+2}}{n+2} = -\log(1-z)-z-\frac{z^2}{2}\tag{1}$$
and
$$ \sum_{n\geq 1}\frac{z^n}{n(n+2)} = \frac{1}{2}\left(-\log(1-z)-\sum_{n\geq 1}\frac{z^n}{n+2}\right)\tag{2}$$
give:
$$ \sum_{n\geq 1}\frac{7n+32}{n(n+2)}\,z^n = \frac{18 z+9 z^2+18 \log(1-z)-32 z^2 \log(1-z)}{2 z^2}.\tag{3} $$
Now just replace $z$ with $\frac{3}{4}$ to see the $\log$-part vanishing and the sum being $\displaystyle\color{red}{\frac{33}{2}}$.
A: Let the $n(\ge1)$th term  $T_n=\dfrac{(7n+32)3^n}{n(n+2)4^n}=A\dfrac{\left(\dfrac34\right)^n}n+B\dfrac{\left(\dfrac34\right)^{n+2}}{n+2}$ where $A,B$ are arbitrary constants
$\implies\dfrac{(7n+32)}{n(n+2)}=\dfrac An+ \dfrac{B\left(\dfrac34\right)^2}{n+2}$ 
$\implies7n+32=A(n+2)+\dfrac{9Bn}{16}$
Comparing the constants, $32=2A\iff A=16$
Comparing the coefficients of $n, 7=A+\dfrac{9B}{16}\iff B=\dfrac{16(7-A)}9=\cdots=-16$
$\implies T_n=16\cdot\dfrac{\left(\dfrac34\right)^n}n-16\cdot\dfrac{\left(\dfrac34\right)^{n+2}}{n+2}=16\left(u_n-u_{n+2}\right)$ 
where $u_m=\dfrac{\left(\dfrac34\right)^m}m$
$\sum_{n=1}^\infty T_n$ clearly telescopes to $16(u_1+u_2)=?$
