Binomial expansion of $(1+x)^{2n}$ and proof that $\binom{2n}{n} = \binom{n}{0}^2 + \binom{n}{1}^2 + \binom{n}{2}^2 + ... + \binom{n}{n}^2$

Question

By considering the coefficient of $x^n$ in the expansion of $(1+x)^{2n}$, show that $\binom{2n}{n} = \binom{n}{0}^2 + \binom{n}{1}^2 + \binom{n}{2}^2 + ... + \binom{n}{n}^2$.

Note: You might find it helpful to use $(1+x)^{2n} = (1+x)^n(1+x)^n$ and that $\binom{n}{n-r} = \binom{n}{r}$.

I have considered expanding it with $(1+x)^{2n} = (1+x)^n(1+x)^n$, however, after expansion I don't know how to approach it.

In addition, I've tried expanding it as follows: $$(1+x)^{2n} = \binom{2n}{0} + \binom{2n}{1}x + \binom{2n}{2}x^2 +...+ \binom{2n}{n}x^n + ... + \binom{2n}{2n}x^{2n}$$ but I don't know how to approach it as well.

Answer 1

We have $$(1+x)^n(1+x)^n=\left({{n}\choose{0}} + {{n}\choose{1}}x+\ldots+{{n}\choose{n}}x^n \right)\left({{n}\choose{0}} + {{n}\choose{1}}x+\ldots+{{n}\choose{n}}x^n \right)$$ The $x^n$ term in this expansion is $$\sum_{k=0}^n{n\choose k}x^k\cdot{n\choose n-k}x^{n-k}=\sum_{k=0}^n{n\choose k}^2x^n$$