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

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.

• From all the answers below, I understand it now. Thank you everyone! – cosmicbrownie Oct 3 '18 at 20:00

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.

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.

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

with $$x = 1$$

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?