What is the value of the folowing sum and why: $$\sum_{n=0}^k\binom{2n}{n}\cdot\binom{2(k-n)}{k-n}$$
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Probably helps to use "Gauss summation" here. Write the series as $\sum a_n$ and pair terms $a_n$ and $a_{k-n}$. – parsiad Oct 16 '20 at 23:19
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There is a (difficult) combinatorial proof here and a couple more proofs, one combinatorial, here. – Brian M. Scott Oct 16 '20 at 23:39
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$$\sum_{k \ge 0} \left(\sum_{n=0}^k a_n b_{k-n}\right) z^k= \left(\sum_{k\ge 0} a_k z^k\right)\left(\sum_{k\ge 0} b_k z^k\right)$$ Take $a_n=b_n=\binom{2n}{n}$ to obtain \begin{align} \sum_{k \ge 0} \left(\sum_{n=0}^k \binom{2n}{n} \binom{2(k-n)}{k-n}\right) z^k &= \left(\sum_{k\ge 0} \binom{2k}{k} z^k\right)\left(\sum_{k\ge 0} \binom{2k}{k} z^k\right) \\ &= \left(\sum_{k\ge 0} \binom{2k}{k}z^k\right)^2 \\ &= \left(\frac{1}{\sqrt{1-4z}}\right)^2 \\ &= \frac{1}{1-4z} \\ &= \sum_{k \ge 0} (4z)^k \\ &= \sum_{k \ge 0} 4^k z^k, \end{align} so $$\sum_{n=0}^k \binom{2n}{n} \binom{2(k-n)}{k-n} = 4^k.$$
RobPratt
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Downvoting a perfectly good answer is misleading, which in my view makes it vandalism. – Brian M. Scott Oct 16 '20 at 23:44
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Agree. It is kind of sad to see -1 when I read the solution. Don't know what kind of person did it. Sick! – toronto hrb Oct 16 '20 at 23:54