Question

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

• 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 $\frac{\left(2n\right)!}{(n-2)!(n+2)!}$

${(1+x)}^n=C_0+C_{1}x+C_2{x}^2+C_3{x}^3+C_4{x}^4+...+C_{n-2}{x}^{n-2}+C_{n-1}{x}^{n-1}+C_n{x}^n$ ...(1)

${(x+1)}^n=C_0{x}^n+C_1{x}^{n-1}+C_2{x}^{n-2}+C_3{x}^{n-3}+C_4{x}^{n-4}+...+C_{n-2}{x}^2+C_{n-1}{x}^1+C_n$ ...(2)

Multiplying (1) and (2) and equating the coefficients of $x^{n-2}$, we get

$C_0{C}_2+C_1{C}_3+C_2{C}_4+...+C_{n-2}{C}_n$

$=$ the coefficients of $x^{n-2}$ in ${(1+x)}^{2n}$

${=}^{2n}{C}_{n-2}=\frac{\left(2n\right)!}{(n-2)!(n+2)!}$