<s>[INST] Hard summation involving binomial and quadratic What is $$\sum \frac{2r^2-98r+1}{(100-r)({100\choose r})}$$  Where $r\in [1,99]$I have reduced it to $$\frac{(2r^2-98r+1)}{(100){99\choose r}}$$ what to do further? Partial fractions don't seem to help.
 [/INST] Using that
$$2r^2-98r+1=(r+1)^2-r(100-r)$$
yields a telescopic sum : 
$$\begin{align}\sum_{r=1}^{99}\frac{2r^2-98r+1}{(100-r)\binom{100}{r}}&=\sum_{r=1}^{99}\frac{(2r^2-98r+1)\cdot r!\cdot (100-r)!}{(100-r)\cdot 100!}\\\\&=\frac{1}{100!}\sum_{r=1}^{99}(2r^2-98r+1)\cdot r!\cdot (99-r)!\\\\&=\frac{1}{100!}\sum_{r=1}^{99}(r^2+2r+1-100r+r^2)\cdot r!\cdot (99-r)!\\\\&=\frac{1}{100!}\sum_{r=1}^{99}((r+1)^2-r(100-r))\cdot r!\cdot (99-r)!\\\\&=\frac{1}{100!}\sum_{r=1}^{99}((r+1)\cdot (r+1)!\cdot (99-r)!-r\cdot r!\cdot (100-r)!)\\\\&=\frac{1}{100!}(100\cdot 100!\cdot 0!-1\cdot 1!\cdot 99!)\\\\&=\color{red}{\frac{9999}{100}}\end{align}$$
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