# 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? Sep 26, 2019 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, 2019 at 15:45
• I get it. Thank you! Sep 26, 2019 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, 2019 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 $x=1.$ Sep 26, 2019 at 16:09

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