Question

If $n\ge 2,$ then $3{\times }^n{C}_1-4{\times }^n{C}_2+5{\times }^n{C}_3-\cdots +(-1{)}^{n-1}(n+2){\times }^n{C}_n=$

• A. $-1$
• B. $2$
• C. $-2$
• D. $1$

Answer 1

Answer: (B)

$2$

We know, $(1-x{)}^n{=}^n{C}_0{-}^n{C}_{1}x{+}^n{C}_2{x}^2-.............+(-1{)}^n$${}^n{C}_n{x}^n$
Now multiply both side $x^2$
$\Rightarrow x^2(1-x{)}^n{=}^n{C}_0{x}^2{-}^n{C}_1{x}^3{+}^n{C}_2{x}^4-.............+(-1{)}^n$${}^n{C}_n{x}^{n+2}$
Differentiating both side w.r.t $x$
$\Rightarrow 2x(1-x{)}^n-n{x}^2(1-x{)}^n=2^n{C}_{0}x-3^n{C}_1{x}^2+4^n{C}_2{x}^3-.............+(-1{)}^n(n+2)$${}^n{C}_n{x}^{n+1}$
Putting $x=1$ we get,
$2^n{C}_0-3^n{C}_1+4^n{C}_2-.............+(-1{)}^n(n+2)$${}^n{C}_n=0$
$\therefore 3{\times }^n{C}_1-4{\times }^n{C}_2+5{\times }^n{C}_3-\cdots +(-1{)}^{n-1}(n+2){\times }^n{C}_n=2{\times }^n{C}_0=2$