Telescoping(?) an infinite series

HINT:

$$7n+32=4^2(n+2)-3^2\cdot n$$

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

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

$$\begin{align} \sum_{n=1}^\infty \frac {7n+32}{n(n+2)}\cdot \left(\frac 34\right)^n &=16\sum_{n=1}^\infty \frac{\left(\frac 34\right)^n}{n} -9\sum_{n=1}^\infty \frac{\left(\frac 34\right)^n}{n+2} \\ &=16\;\;\ln4\;\;\;\;-\;\;9\left(\frac {16}9 \ln4 -\frac {11}6\right)\\ &=\color{red}{\frac{33}2} \end{align}$$

Hint: for $|x|<1$ the power series $f(x):=\sum_{n=1}^{\infty}\frac{1}{n}x^n$ is convergent.

Hence $f'(x)=\sum_{n=1}^{\infty}x^{n-1}=\frac{1}{1-x}$ for $|x|<1$.