If `(1+x)^n=C_0+C_1x+C_2x^2+...+C_n x^n ,` then `C_0C_2+C_1C_3+C_2C_4+...+C_(n-2)C_n=` a. `((2n)!)/((n !)^2)` b. `((2n)!)/((n-1)!(n+1)!)` c. `((2n)!)/((n-2)!(n+2)!)` d. none of these

Options

A. `((2n)!)/((n!)^(2))`
B. `((2n)!)/((n-1)!(n+1)^(!))`
C. `((2n)!)/((n-2)!(n+2)!)`
D. none of these

Answer

C

Explanation

`(1+x)^(n)=C_(0)+.^(n)C_(1)x+.^(n)C_(2)x^(2)+.^(n)C_(3)x^(3)+"....."+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_(n-1)x+C_(n)" "(2)`
Multiplying Eqs. (1) and (2) and equating the coefficient of `x^(n-2)`. we get
`C_(0)C_(2)+C_(1)C_(3)+C_(2)C_(4)+"....."C_(n-2)C_(n)`
`=` coefficient of `x^(n-2)` in `(1+x)^(2n)`
`= .^(2n)C_(n-2)`
`= ((2n)!)/((n-2)!(n+2)!)`