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.
 A: Using that
$$2r^2-98r+1=(r+1)^2-r(100-r)$$
yields a telescopic sum : 
$$\begin{align}\sum_{r=1}^{99}\frac{2r^2-98r+1}{(100-r)\binom{100}{r}}&=\sum_{r=1}^{99}\frac{(2r^2-98r+1)\cdot r!\cdot (100-r)!}{(100-r)\cdot 100!}\\\\&=\frac{1}{100!}\sum_{r=1}^{99}(2r^2-98r+1)\cdot r!\cdot (99-r)!\\\\&=\frac{1}{100!}\sum_{r=1}^{99}(r^2+2r+1-100r+r^2)\cdot r!\cdot (99-r)!\\\\&=\frac{1}{100!}\sum_{r=1}^{99}((r+1)^2-r(100-r))\cdot r!\cdot (99-r)!\\\\&=\frac{1}{100!}\sum_{r=1}^{99}((r+1)\cdot (r+1)!\cdot (99-r)!-r\cdot r!\cdot (100-r)!)\\\\&=\frac{1}{100!}(100\cdot 100!\cdot 0!-1\cdot 1!\cdot 99!)\\\\&=\color{red}{\frac{9999}{100}}\end{align}$$
A: Just to note that, in general $\sum_{r=0}^n \frac{P(r)}{\binom nr}$ can be expressed in terms of $\sum_{r=0}^n \frac1{\binom nr}$ and $n$ only.

Let's denote $c_{np} = \sum_{r=0}^n \frac{r^p}{\binom nr}$. Note that
$$
c_{n1} = \sum_{r=0}^n\frac r{\binom nr} 
= \sum_{r=0}^n\frac{n-r}{\binom n{n-r}} 
= \sum_{r=0}^n\frac{n-r}{\binom nr} 
= nc_{n0}-c_{n1}
\implies c_{n1} = \frac n2c_{n0}
\tag1
$$
similarly we have
$$ c_{n3} = \frac32n c_{n2} - \frac14n^3 c_{n0}. \tag3 $$
From this all odd terms can be written in terms of the even terms. 

These can also be generated using the telescoping sum method from https://math.stackexchange.com/a/1767041, i.e. if we can have the sum
$$ \sum_{r=0}^n \frac{p(r) - q(r+1)}{\binom nr} = p(0) - q(n+1) $$
requiring $\binom n{r-1}p(r)=\binom nrq(r) \implies rp(r) = (n+1-r)q(r)$, i.e. the following equation is satisfied:
$$ \sum_{r=0}^n \frac{(n+1-r)s(r)-(r+1)s(r+1)}{\binom nr} = (n+1)\bigl(s(0) - s(n+1)\bigr) $$
OP's problem is recovered with $s(r)=r$ (and $n=99$), which proves
$$ -2c_{n2} + (n-1)c_{n1} - c_{n0} = -(n+1)^2 
\iff c_{n2} = \frac12(n+1)^2-\frac12(n+1)(n-2)c_{n0} \tag2 $$
Similarly using $s(r)=r^{p-1}$ we could find the relation of $c_{np}$ in terms of the lower terms.

Besides telescoping sum, we could also prove (2) using mathematical induction though it is pretty tedious. We will rely on
\begin{align}
c_{k+1,0} &= \sum_{r=0}^{k+1} \frac1{\binom{k+1}r} \\
&= 1 + \sum_{r=1}^{k+1} \frac1{\binom{k+1}r} \\
&= 1 + \sum_{r=0}^{k} \frac1{\binom{k+1}{r+1}} \\
&= 1 + \sum_{r=0}^{k} \frac1{\frac{k+1}{r+1}\binom kr}\\
&= 1 + \frac{c_{k1} + c_{k0}}{k+1} \\
&= 1 + \frac{\frac k2c_{k0} + c_{k0}}{k+1} \\
\Longleftrightarrow c_{k0} &= \frac{k+1}{\frac12k + 1}(c_{k+1,0}-1), \tag{a}
\end{align}
then
\begin{align}
c_{k+1,2} &= \sum_{r=0}^{k+1} \frac{r^2}{\binom{k+1}r} \\
&= \sum_{r=1}^{k+1} \frac{r^2}{\binom{k+1}r} \\
&= \sum_{r=0}^{k} \frac{(r+1)^2}{\binom{k+1}{r+1}} \\
&= \sum_{r=0}^{k} \frac{(r+1)^2}{\frac{k+1}{r+1}\binom kr} \\
&= \frac1{k+1} \sum_{r=0}^{k} \frac{(r+1)^3}{\binom kr} \\
&= \frac{c_{k3} + 3c_{k2} + 3c_{k1} + c_{k0}}{k+1} \\
&= \frac{(3+\frac32k)c_{k2} + (-\frac14k^3 + \frac32k + 1)c_{k0}}{k+1} &(\text{apply (1), (3)}) \\
&= \frac{(k+2)\bigl(6(k+1)^2 + (k^2+k-2)c_{k0}\bigr)}{8(k+1)} &(\text{apply (2) for }n=k) \\
&= \frac{(k+2)\left(6(k+1)^2 + (k^2+k-2)\frac{k+1}{\frac12k+1}(c_{k+1,0}-1)\right)}{8(k+1)} &(\text{apply (a)}) \\
&= \frac12(k+2)^2 + \frac14(k-1)(k+2)c_{k+1,0} &(\text{simplify yourself ☹})
\end{align}
Thus proves (2) for all $n\ge 0$.

There is no closed form of $c_{n0}$, though from Calculate sums of inverses of binomial coefficients we could eliminate the binomial term:
$$ c_{n0} = \sum_{r=0}^n \frac1{\binom nr} = \frac{n+1}{2^n}\sum_{r=0}^n \frac{2^r}{r+1}.\tag0 $$
