Evaluation of \[ \displaystyle \sum^{10}_{k=0}\binom{20+k}{20}\binom{20-k}{10} \]
I find a way using
finite calculus, maybe not the cleanest or shortest but it works. Using recursively the summation by parts formula
\( \displaystyle{
\sum_{k\geqslant 0}f(k)\Delta g(k)=f(k)g(k)\Big|_{k=0}^{k\to \infty }-\sum_{k\geqslant 0}\Delta f(k) \operatorname{E}g(k)\tag1
} \)
where \( \Delta h(k):=h(k+1)-h(k) \) and \( \operatorname{E}h(k):=h(k+1) \) we find that
\( \displaystyle{
\sum_{k\geqslant 0}\binom{n+k}{n}x^k=\frac{{\color{red}{1}}}{(1-x)^{n+1}}\tag2
} \)
where I set \( f(k):=\binom{n+k}{n} \) and \( \Delta g(k):=x^k \) (so it follows that \( g(k):=\frac{x^k}{x-1} \) is a "finite calculus primitive" of \( \Delta g(k) \)). Now note that \( \sum_{k=0}^{10}\binom{20+k}{20}\binom{20-k}{10} \) is the coefficient of \( x^{10} \) in the expansion of the product
\( \displaystyle{
\left(\sum_{k\geqslant 0}\binom{20+k}{20}x^k\right)\cdot \left(\sum_{k\geqslant 0}\binom{10+k}{10}x^k\right)=\frac{{\color{red}{1}}}{(1-x)^{32}}={\color{red}{\sum_{k\geqslant 0} \binom{31+k}{31}x^k}}\tag3
} \)
where the last equality is a consequence of (2). Therefore, comparing coefficients, from (3) it follows that
\( \displaystyle{
\sum_{k=0}^{10}\binom{20+k}{20}\binom{20-k}{10}=\binom{41}{31}=\binom{41}{10}\tag4
} \)
Correction.
Addition: instead of using summation by parts to get (2) we can notice that
\( \displaystyle{
\binom{n+k}{k}=\binom{n+k}{k}=\frac{(n+k)^{\underline{k}}}{k!}=\frac{(n+1)^{\overline{k}}}{k!}=\frac{(-n-1)^{\underline{k}}}{k!}(-1)^k=\binom{-n-1}{k}(-1) ^{k}\tag5
} \)
Therefore using the binomial identity we have that
\( \displaystyle{
\sum_{k\geqslant 0}\binom{n+k}{n}x^k=\sum_{k\geqslant 0}\binom{-n-1}{k}(-x)^k=(1+(-x))^{-n-1}=\frac1{(1-x)^{n+1}}\tag6
} \)
