Proving a rule of combination [duplicate]

Question

I have this question:

Considering $$(1+x)^n(1+x)^n=(1+x)^{2n}$$ and show that: $${n \choose 0}^2+{n \choose 1}^2+{n\choose 2}^2+....+{n\choose n}^2={2n\choose n}$$

I have attempted so far: $$(1+1)^n={n \choose 0}+{n \choose 1}+{n\choose 2}+....+{n\choose n}$$ $$(1+1)^{2n}={2n \choose 0}+{2n \choose 1}+{2n\choose 2}+....+{2n\choose 2n}$$ $$\left({n \choose 0}+{n \choose 1}+{n\choose 2}+....+{n\choose n}\right)^2={2n \choose 0}+{2n \choose 1}+{2n\choose 2}+....+{2n\choose 2n}$$ and this is the part where I am confused about. Please help in where I got wrong and what I should do to show it in the correct method. The question here is to prove that $${n \choose 0}^2+{n \choose 1}^2+{n\choose 2}^2+....+{n\choose n}^2={2n\choose n}$$

Answer 1

Consider the coefficient of $x^n$ in the both sides of $$(1+x)^n(\color{red}{x+1})^n=(1+x)^{2n},$$ i.e. $$\left(\binom{n}{0}x^n+\binom{n}{1}x^{n-1}+\cdots\right)\left(\binom{n}{0}x^0+\binom{n}{1}x^1+\cdots\right)=\left(\binom{2n}{0}x^{2n}+\binom{2n}{1}x^{2n-1}+\cdots+\binom{2n}{n}x^n+\cdots\right)$$

Answer 2

$(1+x)^n(1+x)^n = \left(\sum_{k=0}^n\binom nk x^k\right)^2$

With Cauchy product we obtain $\sum_{k=0}^{2n}(\sum_{j=0}^k \binom{n}{j}\binom{n}{k-j})x^k$

On the other hand $(1+x)^{2n} = \sum_{k=0}^{2n}\binom{2n}{k}x^k$.

If we compare the coefficients of $x^n$ in both polynomial we get $\binom{2n}{n} = \sum_{j=0}^n \binom{n}{j}\binom{n}{n-j} = \sum_{j=0}^n(\binom{n}{j})^2$ a desired.