Combinatorial coefficients squared If $C_0,C_1,C_2,...,C_n$ are the combinatorial coefficients in the expansion of $(1+x)^n$ then prove that:
$$1C_0^2+3C_1^2+5C_3^2+...+(2n+1)C_n^2=\dfrac{(n+1)(2n)!}{n!n!}=(n+1)\binom {2n}n$$
I am able to compute the linear addition but not the squares of coefficients.
Thanks!

$$\sum_{k=0}^{n}(2k+1)\binom{n}{k}^2=\sum_{k=0}^{n}(2k+1)\binom{n}{k}\binom{n}{n-k}=[x^k]\left[(1+x)^n\sum_{k=0}^{n}\binom{n}{k}(2k+1)x^k\right] $$
but by setting $x=z^2$ we have:
$$\sum_{k=0}^{n}\binom{n}{k}(2k+1)x^k=\frac{d}{dz}\sum_{k=0}^{n}\binom{n}{k}z^{2k+1}=(1+x)^{n-1}(1+(2n+1)x)$$
hence:
$$\sum_{k=0}^{n}(2k+1)\binom{n}{k}^2= [x^n]\left[(1+x)^{2n}+2nx(1+x)^{2n-1}\right]=\binom{2n}{n}+2n\binom{2n-1}{n}$$
and simplifying:
$$\sum_{k=0}^{n}(2k+1)\binom{n}{k}^2=(n+1)\binom{2n}{n} $$
as wanted.
