Question

The sum $\sum _{k=0}^{n-2}{}^n{C}_k\cdot {}^n{C}_{k+2}$ equals

• A. ${}^{2n}{C}_{n+2}$
• B. ${}^{2n}{C}_{n+1}$
• C. ${}^{2n}{C}_{n-1}$
• D. ${}^{2n}{C}_n$

Answer 1

Answer: (A)

Write the expansion of $(1+x{)}^n$ in two ways, multiply and compare the co-efficient of $x^{n+2}$.
$S=C_0{C}_2+C_1{C}_3+C_2{C}_4+\dots \dots \dots ..+C_{n-2}{C}_n$
now, $(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+\dots \dots \dots \dots ..+C_n{x}^n\dots \dots \dots ...(1)$
$(x+1{)}^n=C_0{x}^n+C_1{x}^{n-1}+C_2{x}^{n-2}+\dots \dots \dots \dots +C_n$.....(2)
Multiplying (1) and (2) \& equating the coefficient of $x^{n+2}$
$C_0{C}_2+C_1{C}_3+C_2{C}_4+\dots \dots \dots \dots \dots ...coefficient\text{ of }{x}^{n+2}\text{ in }(1+x{)}^{2n}={}^{2n}{C}_{n+2}$