<s>[INST] Evaluate $\sum_{r =1}^{n -1}{\frac{n(n+1)-r(2r+3)-1}{(r+1)^2}\cdot{}^{n+r}{C_r}\cdot {}^n{C_r}}$ 
Evaluate $\sum_{r =1}^{n -1}{\frac{n(n+1)-r(2r+3)-1}{(r+1)^2}\cdot{}^{n+r}{C_r}\cdot {}^n{C_r}}$

I noted that
(A) ${}^{20}{C_{10}} - 110$, if $n=10$
(B) ${}^{22}{C_{11}} - 110$, if $n=11$
(C) ${}^{18}{C_{9}} - 100$, if $n=10$
(D) ${}^{20}{C_{10}} - 121$, if $n=11$
My basic approach is as follows:
$\sum\limits_{r = 1}^{n - 1} {\frac{{n\left( {n + 1} \right) - r\left( {2r + 3} \right) - 1}}{{{{\left( {r + 1} \right)}^2}}}.{}^{n + r}{C_r}.{}^n{C_r}}  = $
$T = \frac{1}{{{{\left( {r + 1} \right)}^2}}}.{}^{n + r}{C_r}.{}^n{C_r}$
$T = \frac{1}{{{{\left( {r + 1} \right)}^2}}}.\frac{{\left( {n + r} \right)!}}{{r!.n!}}.\frac{{n!}}{{r!\left( {n - r} \right)!}} \Rightarrow T = \left( {\frac{1}{{n + r + 1}} - \frac{1}{{n + r + 2}}} \right) \times \frac{{\left( {n + r + 2} \right)\left( {n + r + 1} \right)\left( {n + r} \right)!}}{{\left( {n - r} \right)!\left( {2r + 2} \right)!}}.\frac{{2r + 2!}}{{r + 1!.r + 1!}}$
$T = \left( {\frac{1}{{n + r + 1}} - \frac{1}{{n + r + 2}}} \right) \times \frac{{\left( {n + r + 2} \right)!}}{{\left( {n - r} \right)!\left( {2r + 2} \right)!}}.\frac{{2r + 2!}}{{r + 1!.r + 1!}} \Rightarrow T = \left( {\frac{1}{{n + r + 1}} - \frac{1}{{n + r + 2}}} \right) \times {}^{n + r + 2}{C_{2r + 2}}.{}^{2r + 2}{C_{r + 1}}$
Not able to proceed.
 [/INST] Note that
$$\begin{align}\frac{n(n+1)-r(2r+3)-1}{(r+1)^2}&=\frac{n^2+n-2r^2-3r-1}{(r+1)^2}
\\&=\frac{n^2-r^2+n-r-(r^2+2r+1)}{(r+1)^2}\\
&=\frac{(n+r+1)(n-r)}{(r+1)^2}-1\end{align}$$
which implies
$$\begin{align}\frac{n(n+1)-r(2r+3)-1}{(r+1)^2}&\binom{n+r}{r}\binom{n}{r}\\
&=\frac{n+r+1}{r+1}\binom{n+r}{r}\frac{n-r}{r+1}\binom{n}{r}-\binom{n+r}{r}\binom{n}{r}\\
&=\binom{n+r+1}{r+1}\binom{n}{r+1}-\binom{n+r}{r}\binom{n}{r}\\
&=b_{r+1}-{b_r}\end{align}$$
where $b_r=\binom{n+r}{r}\binom{n}{r}.$
Therefore the sum turns out to be telescopic:
$$\begin{align}\sum_{r=1}^{n-1}\frac{n(n+1)-r(2r+3)-1}{(r+1)^2}\binom{n+r}{r}\binom{n}{r}&=\sum_{r=1}^{n-1}(b_{r+1}-{b_r})\\
&=b_{n}-b_1=\binom{2n}{n}-(n+1)n.
\end{align}$$
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