Question

If $(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+C_3{x}^3+\cdots +C_n{x}^n,$ then $C_0{C}_2+C_1{C}_3+C_2{C}_4+\cdots +C_{n-2}{C}_n=$

• A. $\frac{\left(2n\right)!}{(n!{)}^2}$
• B. $\frac{\left(2n\right)!}{(n-1)!(n+1)!}$
• C. $\frac{\left(2n\right)!}{(n-2)!(n+2)!}$
• D. None of these

Answer 1

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

We know that,
$(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+C_3{x}^3+\cdots +C_n{x}^n\cdots (1)$
$(x+1{)}^n=C_0{x}^n+C_1{x}^{n-1}+C_2{x}^{n-2}+C_3{x}^{n-3}+\cdots +C_n\cdots (2)$

Multiplying $\left(1\right)$ and $\left(2\right)$ and equating the coefficient of $x^{n-2},$ we get
$C_0{C}_2+C_1{C}_3+C_2{C}_4+\cdots +C_{n-2}{C}_n=\text{coefficient of}{x}^{n-2}\text{in}(1+x{)}^{2n}={}^{2n}{C}_{n-2}=\frac{\left(2n\right)!}{(n-2)!(n+2)!}$