Question

If $(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+......+C_n{x}^n$, prove that
$C_0{C}_r+C_1{C}_{r+1}+......+C_{n-r}{C}_n=\frac{\left(2n\right)!}{(n+r)!(n-r)!}$

Answer 1

We have,

${(1+x)}^n=C_0+C_{1}x+C_2{x}^2+...+C_n{x}^n$

${(1+\frac{1}{x})}^n=C_0+C_1\frac{1}{x}+C_2{\left(\frac{1}{x}\right)}^2+...+C_n{\left(\frac{1}{x}\right)}^n$

Given summation$=$Co-efficient of $x^r$ in ${(1+x)}^n{(1+\frac{1}{x})}^n$

$=$Coefficent of $x^r$ in $\frac{{(1+x)}^{2n}}{x^n}$

$=$Coefficent of $x^{n+r}$ in ${(1+x)}^{2n}$

$=\frac{2n!}{(n+r)!(2n-n-r)!}$

$=\frac{2n!}{(n+r)!(n-r)!}$

Hence, proved.