If $C_0, C_1, C_2,$$.............C_n$are the coefficients in the expansion of ${\left(1+x\right)}^n$ then show that $C_{0}Cr+C_1{C}_{r+1}+C_2{C}_{r+2}+\mathrm{.......}+C_{n-r}.C_n=\frac{\left(2n\right)!}{(n-r)!(n+r)!}$


 

$Proof:{(1+x)}^n=C_0+C_{1}x+C_2{x}^2+\mathrm{.......}+C_r{x}^r+$$C_{r+1}{x}^{r+1}+C_{r+2}{x}^{r+2}+\mathrm{.....}+C_{n-r}{x}^{n-r}+\mathrm{........}+C_n{x}^n\mathrm{........}\left(1\right)$

Also it may be written as 
${(x+1)}^n=C_0{x}^n+C_1{x}^{n-1}+C_2{x}^{n-2}+\mathrm{...}+C_{n-r}{x}^r+\mathrm{.....}+C_n$$..........\left(2\right)$

Multiplying the expansion (1) and (2) and equating the coefficients of $x^{n+r}$on both sides, we get 
$C_0{C}_r+C_1{C}_{r+1}+C_2{C}_{r+2}+\mathrm{.....}+C_{n-r}{C}_n={(1+x)}^{2n}$

coefficient $x^{n+r}in{(1+x)}^{2n}$

But in ${(1+x)}^{2n},x^{n+r}$ occurs in ${\left(n+r+1\right)}^{th}$term

$\therefore T_{n+r+1}{=}^{2n}{C}_{n+r}=\frac{\left(2n\right)!}{(n+r)!(n-r)!}{x}^{n+r}$

$C_0{C}_r+C_1{C}_{r+1}+C_2{C}_{r+2}+\mathrm{.......}+C_{n-r}.C_n=\frac{\left(2n\right)!}{(n-r)!(n+r)!}$

Note 1 : If we put r = 1 in the above result i.e., collect the coefficients of $x^{n+1}$in the product of the two expansions, we get 

$C_0{C}_1+C_1{C}_2+C_2{C}_3+\mathrm{.......}+C_{n-1}{C}_n=\frac{\left(2n\right)!}{(n-r)!(n+r)!}$

Note 2  :If we put r = 0 in the above result i.e., collect the coefficients of $x_n$ only in the product of the two expansions, we get 

$C_0^2+C_1^2+C_2^2+\mathrm{........}+C_n^2=\frac{\left(2n\right)!}{{(n!)}^2}$