Question

The value of $(n+2)C_0{2}^{n+1}-(n+1)C_1{2}^n+n{C}_2{2}^{n-1}+....$ is equal to:
$(C_r={}^n{C}_r)$

• A. $4n$
• B. $4$
• C. $2n+4$
• D. $4+4n$

Answer 1

The correct option is D $4+4n$

We know, ${(1-x)}^n{=}^n{C}_0{x}^n{-}^n{C}_1{x}^{n-1}{+}^n{C}_2{x}^{n-2}-...{.}^n{C}_n$

Multiplying by $x^2$ on both sides,

$x^2{(1-x)}^n{=}^n{C}_0{x}^{n+2}{-}^n{C}_1{x}^{n+1}{+}^n{C}_2{x}^n-....$

Taking the derivative,

$2x{(1-x)}^n+n{x}^2{(1-x)}^{n-1}=(n+2)C_0{x}^{n+1}-(n+1)C_1{x}^n+n{C}_2{x}^{n-1}-....$

Putting $x=2$ in LHS, we get our required result,

$4(-1{)}^n+n(4\left)\right(-1{)}^{n-1}$

$=4-4n$, $n$ is even

$=4n-4$, $n$ is odd.

There is $no$ $option$ matching the answer.