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Show $\sum_{k=0}^{n} {2k \choose k}{2(n-k) \choose n-k} = 4^n$.

I use the fact ${2n \choose n} = (-4)^n {- \frac{1}{2} \choose n} = (4)^n {n- \frac{1}{2} \choose n}$ to reduce the sum to $4^n \sum_{k=0}^{n} {k- \frac{1}{2} \choose k} {n-k- \frac{1}{2} \choose n-k}$ which is equal to $4^n {n-1 \choose n}$ using vandermonde's identity which states ${a+b \choose n} = \sum_{k=0}^{n} {a \choose k}{b \choose n-k}$.

Where am I going wrong?

Jakob Streipel
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Ray
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1 Answers1

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We have to avoid $k$ in the top argument when using Vandermonde. (See counterexample below.)

Hence if we do, then
$$ \sum_{k = 0}^n \binom{2k}{k} \binom{2(n - k)}{n - k} = \sum_{k = 0}^n (-4)^k \binom{-1/2}{k} (-4)^{n - k} \binom{-1/2}{n - k} = (-4)^n \sum_{k = 0}^n \binom{-1/2}{k} \binom{-1/2}{n - k},$$ we get, from Vandermonde, $$ \sum_{k = 0}^n \binom{2k}{k} \binom{2(n - k)}{n - k}= (-4)^n \binom{-1}{n} = (-4)^n (-1)^n = 4^n. $$

Note on having $k$ in the top argument: consider the trivial example $$ \sum_{k = 0}^n \binom{k}{k} \binom{n - k}{n - k} = n + 1; $$ attempting to apply Vandermonde here would suggest it is equal to $\binom{n}{n} = 1$.

amWhy
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Jakob Streipel
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  • @amWhy I know the same identity was derived elsewhere---I linked a dupe myself above. I answered because OP had a question about this particular approach, not about any other derivation. – Jakob Streipel Nov 28 '21 at 20:18
  • Thanks for the added info, @prets. I'll delete my first comment! I made mini-edits to enable me to change my vote, as well. – amWhy Nov 28 '21 at 20:22
  • No worries @amWhy. In hindsight the question could maybe be improved by being renamed, since it's really a question about Vandermonde's identity (and how using it carelessly is dangerous), not about the combinatorial identity. – Jakob Streipel Nov 28 '21 at 20:24
  • I agree. And thanks for understanding. ;D – amWhy Nov 28 '21 at 20:27