Equation involving $4^n$

Asked Sep 30 '20 at 18:16

Active May 27 '21 at 19:25

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Show using an algebraic and combinatorial approach that for all $n\geq N,$ $$\sum_{j=0}^n {2j\choose j}{2n-2j\choose n-j}= 4^n.$$

The algebraic approach is not too hard to do using the fact that $\dfrac{1}{\sqrt{1-4x}}=\sum_{k=0}^\infty {2k\choose k}x^k,$ and squaring both sides and taking the coefficient of $x^n$ of both sides. However, I am lost as to how to use the combinatorial approach. I know that ${n\choose k}$ is the number of ways to choose $k$ objects from a group of $n$ objects, but I'm not sure how to apply that to this problem. Also, $4^n$ is the number of ways to choose four types of object in $n$ draws with replacement.

edited May 27 '21 at 19:25

MarianD

2,953

asked Sep 30 '20 at 18:16

Take a look here: https://math.stackexchange.com/questions/72367/proof-of-a-combinatorial-identity-sum-limits-i-0n-2i-choose-i2n-i-ch – Phicar Sep 30 '20 at 18:18
And Marc van Leeuwen gives a different combinatorial proof here. – Brian M. Scott Sep 30 '20 at 18:20
The question seems incomplete. Show that $\sum_{j=0}^n {2j \choose j}{2n - 2j \choose n - j}$ is equal to what? – Robby the Belgian Sep 30 '20 at 18:21
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Equal to $4^n$ is missing. – Dietrich Burde Sep 30 '20 at 18:21
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Does this answer your question? Identity for convolution of central binomial coefficients: $\sum\limits_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}$ – Dietrich Burde Sep 30 '20 at 18:21

Equation involving $4^n$

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