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.

HINT:
$$7n+32=4^2(n+2)-3^2\cdot n$$
$$\implies\dfrac{7n+32}{n(n+2)}\left(\dfrac34\right)^n=\dfrac{4^2(n+2)-3^2\cdot n}{n(n+2)}\left(\dfrac34\right)^n=\dfrac{16\left(\dfrac34\right)^n}n-\dfrac{16\left(\dfrac34\right)^{n+2}}{(n+2)}$$
