Question

If ${\left(1+x\right)}^n=C_0+C_{1}x+C_2{x}^2+.......+C_n{x}^n$. then $C_0+2C_1+3C_2+.........+\left(n+1\right)C_n$ is equal to-

• A. $2^{n-1}\left(n+2\right)$
• B. $2^n\left(n+1\right)$
• C. $2^{n-1}\left(n+1\right)$
• D. $2^n\left(n+2\right)$

Answer 1

The correct option is A $2^{n-1}\left(n+2\right)$

$(Hn{)}^n+C_0+C_{1}x+C_2{x}^2+........C_n{x}^n$

$C_0+2C_1+3C_2+.........+(n+1)C_n=2$

$(Hn{)}^n=C_0+C_{1}x+C_2{x}^2+............C_n{x}^n-----(1)$

$n(1+x{)}^{n-1}=0+C_1+2C_{2}x+3C_3{x}^2+.....+x{C}_n{x}^{n-1}$

Put $x=1$

$n(2{)}^{n-1}=C_1+2C_2+3C_3+.......+n{C}_n--------(2)$

Put $x=1$ in $\left(1\right)$

$2^n=C_0+C_1+C_2+C_3+.......C_n---------\left(3\right)$

$\left(2\right)+\left(3\right)$ & $C_0+2C_1+3C_2+..........+(n+1)C_n=n{2}^{n-1}+2^n$

$=2^n[1+\frac{n}{2}]$

$C_0+2C_1+3C_2+........(n+1)C_n=2^{n+1}[2+n]$