proving $\displaystyle\sum_{s,r=0,s\ne r}^{n}C_rC_s=\frac{1}{2}\left(2^{2n}-\frac{(2n)!}{(n!)^2}\right)$

Question

Using $\displaystyle(1+x)^n=\sum_{r=0}^n C_rx^r$, find $\displaystyle\sum_{r=0}^n C_r$ and $\displaystyle\sum_{r=0}^n C_r^2$. Hence prove sum of product of two coefficients taken from $C_0,C_1,C_2\cdots,C_n$ is $\displaystyle\frac{1}{2}\left(2^{2n}-\frac{(2n)!}{(n!)^2}\right)$. i.e. $\displaystyle\sum_{s,r=0,s\ne r}^{n}C_rC_s$

Here's what I tried so far,

$$\sum_{r=0}^n C_r=2^n$$

using the fact that $(1+x)^{2n}$, I was able to derive,

$$\sum_{r=0}^n C_r^2=\frac{(2n)!}{(n!)^2}$$

But How can I proceed? Any Hint would be highly appreciated. Thank you.

Answer 1

$\sum_{s,r=0.s\ne r}^nC_rC_s=(\sum_{r=0}^nC_r)^2-\sum_{r=0}^nC_r^2=2^{2n}-\frac{(2n)!}{(n!)^2}$ To get the $\frac{1}{2}$ you need to change $s\ne r$ to $s\lt r$.

Answer 2

In general for any sequence of real numbers $a_1,a_2,\cdots,a_n$, we have that $$2\sum_{1 \le i <j \le n} a_i a_j=\left(\sum_{1 \le i \le n} a_i\right)^2-\left(\sum_{1 \le i \le n} a_i^2\right)$$ and this property is used here.