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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1Well, 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? – PhoemueX Sep 26 '19 at 15:39
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Binomial Theorem: $$(1+x)^n=\sum_{k=0}^n x^k \binom nk$$ look at what do you get when you set $x=1$ – WW1 Sep 26 '19 at 15:45
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I get it. Thank you! – Wang Chen Sep 26 '19 at 15:49
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Not to be confused with another identity, $\sum_{k=0}^n\binom{n}{k}^2=\binom{2n}{n}$. – J.G. Sep 26 '19 at 15:52
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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 $x=1.$ – Thomas Andrews Sep 26 '19 at 16:09
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Possible duplicate of Non-inductive, not combinatorial proof of $\sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$ – lab bhattacharjee Sep 26 '19 at 16:22
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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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