Question

If $(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+.......+C_n{x}^n$, then the value of $C_0{C}_r+C_1{C}_{r+1}+C_2{C}_{r+2}+...C_{n-r}{C}_n=$

• A. $\frac{\left(2n\right)!}{(n-r)!(n+r)!}$
• B. $\frac{n!}{(-r)!(n+r)!}$
• C. $\frac{n!}{(n-r)!}$
• D. none of these

Answer 1

The correct option is A $\frac{\left(2n\right)!}{(n-r)!(n+r)!}$

Consider the following series.
${}^m{C}_0{}^n{C}_r+{}^m{C}_1{}^n{C}_{r-1}+...{}^m{C}_r{}^n{C}_0$
=Coefficient of $x^r$ in $(1+x{)}^m.(x+1{)}^n$
=coefficient of $x^r$ in $(1+x{)}^{m+n}$
$={}^{m+n}{C}_r$
In the above case,
$r^{\prime }=n-r$ and $m=n$
Hence
${}^{2n}{C}_{n-r}$
$=\frac{2n!}{(n+r)!(n-r)!}$