Show that $[\sum\limits_{k = 0}^n\binom{n}{k}]^2=\sum\limits_{r = 0}^{2n}\binom{2n}{r}$

Question

I'm not sure how to show that $[\sum\limits_{k = 0}^n\binom{n}{k}]^2=\sum\limits_{r = 0}^{2n}\binom{2n}{r}$. I've heard of $\sum\limits_{k = 0}^n {n \choose k}^2= {2n \choose n}$ but I still get nowhere. I have no idea where $r$ came from or why it's $2n$ above the sum.

Answer 1

By using a basis propertie of Pascal's triangle the LHS can be written as

$$\left[\sum_{k=0}^n\binom{n}{k}\right]^2=[2^n]^2=2^{2n}=\sum_{r=0}^{2n}\binom{2n}{r}$$

The $r$ is just used as a new index to avoid confusions by naming two different variables $k$ for example.

Answer 2

Use $$\sum_{f=0}^g{g\choose f}=2^g$$ substituting appropriate values for $f,g$ on both sides of what you want to prove.

Answer 3

You could also do a combinatorial proof: given $n$ distinct dogs and $n$ distinct cats, how many ways can you choose a subset of all the animals to take for a walk?

Answer 4

$$ \sum_{k=0}^n\binom{n}{r}x^r = (x+1)^n\Rightarrow (x+1)^{2n} = \sum_{k=0}^{2n}\binom{2n}{r}x^r $$

with $x = 1$