Question

If $C_0,C_1,C_2,.......C_n$ denote the coefficients in the expansion of ${(1+x)}^n$, prove that
$C_1+2C_2+3C_3+....n{C}_n=n.{\left(2\right)}^{n-1}$

Answer 1

If $C_0,C_1,C_2,C_3...$ denote the coefficients in the expansion of $(1+x{)}^n$

$\Rightarrow (1+x{)}^n=C_0+C_{1}x+C_2{x}^2+C_3{x}^3+C_4{x}^4+C_5{x}^5+C_6{x}^6+....C_n{x}^n$

On differentiating both sides with respect to x, we get

$\Rightarrow n(1+x{)}^{n-1}=C_1+2C_{2}x+3C_3{x}^2+4C_4{x}^3+5C_5{x}^4+6C_6{x}^5+7C_7{x}^6+....n{C}_0{x}^{n-1}$

On putting $x=1$,we get the relation

$n{.2}^{n-1}=C_1+2C_2+3C_3+4C_4+5C_5+6C_6+7C_7+....n{C}_n$

Hence, proved.