Stanford 2011

$\begin{aligned} S & = \sum_{n=1}^\infty \frac {7n+32}{n(n+2)}\left(\frac 34\right)^n \\ & = \sum_{n=1}^\infty \left(\frac 7n + \frac {18}{n(n+2)}\right)\left(\frac 34\right)^n \\ & = \sum_{n=1}^\infty \left(\frac 7n + 9 \left(\frac 1n - \frac 1{n+2} \right)\right)\left(\frac 34\right)^n \\ & = \sum_{n=1}^\infty \left(\frac {16}n\left(\frac 34\right)^n - \frac 9{n+2} \left(\frac 34\right)^n \right) \\ & = \sum_{n=1}^\infty \left(\frac {16}n\left(\frac 34\right)^n - \frac 9{n+2} \times \frac {16}9 \times \left(\frac 34\right)^{n+2} \right) \\ & = 16 \sum_{n=1}^\infty \left(\frac 1n \left(\frac 34\right)^n - \frac 1{n+2} \left(\frac 34\right)^{n+2} \right) \\ & = 16 \left(\sum_{n=1}^\infty \frac 1n \left(\frac 34\right)^n - \sum_{n=3}^\infty \frac 1n \left(\frac 34\right)^n \right) \\ & = 16 \left(\frac 11\times \frac 34 + \frac 12 \times \frac 9{16} \right) \\ & = \frac {33}2 = \boxed{16.5} \end{aligned}$

Similar to another solution, but this can easily be done using partial fraction decomposition: $\frac{7n+32}{n(n+2)}=\frac{16}n-\frac9{n+2}.$ Now, since $9=(3/4)^2\cdot 16,$ we get $\sum_{n=1}^\infty \frac{16}n\left(\frac34\right)^n-\frac{16}{n+2}\left(\frac34\right)^{n+2}$ and the answer follows.

Using partial fractions, we find that the summand can be written as (16/n) (x^n) - (9/(n+2)) x^n, where x = 3/4. Now use the identity, for 0<x<1, - ln(1-x) = (sum(x^n)/n from n = 1 to n = infinity) (NOTE: This identity can be obtained by integrating the geometric series term by term). Most of the terms in the sum cancel, with the exception of 16 x + (16/2) x^2. Substitute x = 3/4 to find this to get the required result.