Telescoping(?) an infinite series Find the value of the sum $\displaystyle \sum\limits_{n=1}^{\infty} \frac{(7n+32) \cdot3^n}{n(n+2) \cdot 4^n}.$
Using partial fraction decomposition, I found the above expression is equivalent to $\displaystyle \sum\limits_{n=1}^{\infty} \frac{25}{n} \cdot \left(\frac{3}{4}\right)^n - \sum\limits_{n=1}^{\infty} \frac{18}{n+2} \cdot \left(\frac{3}{4}\right)^n,$ where I got couldn't find a closed form of either expression because of the $\left(\dfrac{3}{4}\right)^n.$
Similarly, trying to telescope one of the the sums with $\displaystyle \sum\limits_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+2}\right) \cdot \left(\frac{3}{4}\right)^n$ fails for the same reason. How can I further simplify the above expression? Thanks.

Note that
$$\sum_{n=1}^\infty \frac {r^n}n=\sum_{n=1}^\infty \int_0^x r^{n-1}dr=\int_0^x\sum_{n=1}^\infty r^n dr=\int_0^x \frac 1{1-r}dr=\bigg[-\ln(1-r)\bigg]_0^x=\ln\left(\frac 1{1-x}\right)$$
and
$$\sum_{n=1}^\infty\frac {r^n}{n+2}=\sum_{n=3}^\infty \frac {r^{n-2}}{n}=\frac 1{r^2}\sum_{n=3}^\infty\frac {r^n}n=\frac 1{r^2}\left[\left(\sum_{n=1}^\infty\frac {r^n}n\right)-r-\frac {r^2}2\right]=\frac 1{r^2}\left[\ln\left(\frac 1{1+r}\right)\right]-\frac 1r-\frac 12$$
Putting $r=\dfrac 34$ gives
$$\sum_{n=1}^\infty \frac{\left(\frac 34\right)^n}n=\ln 4$$
and $$\sum_{n=1}^\infty \frac {\left(\frac 34\right)^n}{n+2}=\frac {16}9\ln 4-\frac {11}6$$
Hence 
$$\begin{align}
\sum_{n=1}^\infty \frac {7n+32}{n(n+2)}\cdot \left(\frac 34\right)^n
&=16\sum_{n=1}^\infty \frac{\left(\frac 34\right)^n}{n} 
-9\sum_{n=1}^\infty \frac{\left(\frac 34\right)^n}{n+2} \\
&=16\;\;\ln4\;\;\;\;-\;\;9\left(\frac {16}9 \ln4 -\frac {11}6\right)\\
&=\color{red}{\frac{33}2}
\end{align}$$
