Question

If $(1+x{)}^{15}=C_0+C_{1}x+C_2{x}^2+\dots \dots +C_{15}{x}^{15},$ find the value of $C_2+2C_3+3C_4+\dots \dots +14C_{15}$

Answer 1

${(1+x)}^{15}=C_0+C_{1}x+C_2{x}^2...........+C_{15}{x}^{15}$

so, $C_2+2C_3+......14C_{15}$ is the expansion of $\frac{d}{dx}\left(\frac{{(1+x)}^{15}-1}{x}\right)$ at $x=1$

so, $\frac{d}{dx}\left(\frac{{(1+x)}^{15}-1}{x}\right)=\frac{x\left(15{(1+x)}^{14}\right)-({(1+x)}^{15}-1)}{x^2}$

at $x=1,$ this value tends to

$\Rightarrow \frac{1\left(15{\left(2\right)}^{14}\right)-(2^{15}-1)}{1}$

$\Rightarrow \frac{2^{14}\left(13\right)+1}{1}$

so, $C_2+2C_3+3C_4+..........+14C_{15}=\frac{2^{14}\left(13\right)+1}{1}$

Hence, the answer is $\frac{2^{14}\left(13\right)+1}{1}.$