How to show that $$\left[\sum_{k=0}^n \binom{n}{k}\right]^2 = \sum_{k=0}^{2n} \binom{2n}{k}$$

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    $\begingroup$ Well, you can explicitly compute both the left- and the right side. Then the identity should be obvious. Also, what properties of the binomial coefficients do you know? What are your ideas? Where are you stuck? $\endgroup$
    – PhoemueX
    Sep 26, 2019 at 15:39
  • $\begingroup$ Binomial Theorem: $$(1+x)^n=\sum_{k=0}^n x^k \binom nk$$ look at what do you get when you set $x=1$ $\endgroup$
    – WW1
    Sep 26, 2019 at 15:45
  • $\begingroup$ I get it. Thank you! $\endgroup$
    – Wang Chen
    Sep 26, 2019 at 15:49
  • $\begingroup$ Not to be confused with another identity, $\sum_{k=0}^n\binom{n}{k}^2=\binom{2n}{n}$. $\endgroup$
    – J.G.
    Sep 26, 2019 at 15:52
  • $\begingroup$ You have more generally $$\left[\sum_{k=0}^n \binom{n}{k}x^k\right]^2 = \left[(1+x)^n\right]^2=(1+x)^{2n}=\sum_{k=0}^{2n} \binom{2n}{k}x^{k}$$ Your case is just $x=1.$ $\endgroup$ Sep 26, 2019 at 16:09

1 Answer 1


The easiest way, assuming you know that $\sum_{k=0}^n \binom{n}{k}=2^n$, is doing the following: $$\left[\sum_{k=0}^n \binom{n}{k}\right]^2 =(2^n)^2=2^{2n}= \sum_{k=0}^{2n} \binom{2n}{k}. \qquad QED$$


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