Bashing Telescope

In the true spirit of bashing, we hope to find the partial fraction form of the expression.

By inspection (rational root theorem helps too), observe that
$n^5+10n^4+35n^3+50n^2+24n=n(n+1)(n+2)(n+3)(n+4)$

Hence let $\frac {7n^3+41n^2+88n+72}{n^5+10n^4+35n^3+50n^2+24n}=\frac {A}{n}+ \frac {B}{n+1} +\frac {C}{n+2} +\frac {D}{n+3}+ \frac {E}{n+4}$

Multiplying by the original denominator on both sides, $7n^3+41n^2+88n+72=$ $A(n+1)(n+2)(n+3)(n+4)$ $+Bn(n+2)(n+3)(n+4)$ $+Cn(n+1)(n+3)(n+4)$ $+Dn(n+1)(n+2)(n+4)$ $+En(n+1)(n+2)(n+3)$

Now, the easiest way is to set $n=0,-1,-2,-3,-4$ to get a linear equation in terms of each coefficient A,B,C,D and E respectively.

Note: If this were to be a formal proof, we would have to expand and compare coefficients, since by right n cannot be equal to -4,-3,-2,-1,0 otherwise we would have division by zero. Personally I feel this informal method is much faster.

We then obtain $(A,B,C,D,E)=(3,-3,1,2,-3)$

$\frac {7n^3+41n^2+88n+72}{n^5+10n^4+35n^3+50n^2+24n}=\frac {3}{n}- \frac {3}{n+1} +\frac {1}{n+2} +\frac {2}{n+3}- \frac {3}{n+4}$

Finally $\displaystyle \sum_{n=1}^{24}\frac {7n^3+41n^2+88n+72}{n^5+10n^4+35n^3+50n^2+24n}$ $=\displaystyle \sum_{n=1}^{24} (\frac {3}{n}- \frac {3}{n+1} +\frac {1}{n+2} - \frac {1}{n+4}+\frac {2}{n+3}- \frac {2}{n+4})$ $= 3\displaystyle \sum_{n=1}^{24} (\frac {1}{n}-\frac {1}{n+1})+ \displaystyle \sum_{n=3}^{26} (\frac {1}{n}-\frac {1}{n+2})+ 2\displaystyle \sum_{n=4}^{27} (\frac {1}{n}-\frac {1}{n+1})$ $=3(1-\frac {1}{25})+(\frac {1}{3}+ \frac {1}{4}- \frac {1}{27} -\frac {1}{28})+2 (\frac {1}{4}- \frac {1}{28})$ $=\frac {36091}{9450}$ $\Rightarrow A+B=\boxed{45541}$

Let the sum be $S$ , then we have:

$\begin{aligned} S & = \sum_{n=1}^{24} \frac{7n^3+41n^2+88n+72}{n^5+10 n^4 + 35 n^3+50n^2+24n} \\ & = \sum_{n=1}^{24} \frac{3(n^3+9n^2+26n+24)+4n(n^2+3n+2)+ 2n(n+1)}{n(n+1)(n+2)(n+3)(n+4)} \\ & = \sum_{n=1}^{24} \frac{3(n+2)(n+3)(n+4)+4n(n+1)(n+2)+2n(n+1)}{n(n+1)(n+2)(n+3)(n+4)} \\ & = \sum_{n=1}^{24} \left(\frac{3}{n(n+1)} + \frac{4}{(n+3)(n+4)} + \frac{2}{(n+2)(n+3)(n+4)} \right) \\ & = \sum_{n=1}^{24} \left(3 \left[ \frac{1}{n} - \frac{1}{n+1} \right] + 4 \left[ \frac{1}{n+3} - \frac{1}{n+4} \right] + \left[ \frac{1}{n+2} - \frac{2}{n+3} + \frac{1}{n+4} \right] \right) \\ & = 3 \sum_{n=1}^{24} \frac{1}{n} - 3 \sum_{n=1}^{24} \frac{1}{n+1} + \sum_{n=1}^{24} \frac{1}{n+2} + 2 \sum_{n=1}^{24} \frac{1}{n+3} - 3 \sum_{n=1}^{24} \frac{1}{n+4} \\ & = 3 \sum_{n=1}^{24} \frac{1}{n} - 3 \sum_{n=2}^{25} \frac{1}{n} + \sum_{n=3}^{26} \frac{1}{n} + 2 \sum_{n=4}^{27} \frac{1}{n} - 3 \sum_{n=5}^{28} \frac{1}{n} \\ & = \frac{3}{1} - \frac{3}{25} + \frac{1}{3} + \frac{3}{4} - \frac{1}{27} - \frac{3}{28} \\ & = \frac{74}{25} + \frac{8}{27} + \frac{9}{14} \\ & = \frac{27216+2800+6075}{9450} = \frac{36091}{9450} \end{aligned}$

$\Rightarrow A+B = 36091+9450 = \boxed{45541}$