# Prove that $\left[\sum_{k=0}^n \binom{n}{k}\right]^2 = \sum_{k=0}^{2n} \binom{2n}{k}$ [closed]

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

• 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? – PhoemueX Sep 26 '19 at 15:39
• 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
• I get it. Thank you! – Wang Chen Sep 26 '19 at 15:49
• 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
• 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

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