Question

Prove that $C_0+2\cdot C_1+3\cdot C_2+.....+(n+1)C_n=2^{n-1}\cdot (n+2)$

Answer 1

We have ${(1+x)}^n=c_0+c_{1}x+c_2{x}^2+c_3{x}^3+...+c_n{x}^n$ .....$\left(1\right)$

Multiply $\left(1\right)$ with $x$ we get

${(1+x)}^{n}x=c_{0}x+c_1{x}^2+c_2{x}^3+c_3{x}^4+...+c_n{x}^{n+1}$ .......$\left(2\right)$

Differentiating $\left(2\right)$ w.r.t $x$ we get

${(1+x)}^n+nx{(1+x)}^{n-1}=c_0+2c_{1}x+3c_2{x}^2+4c_3{x}^3+...+(n+1)c_n{x}^n$ .......$\left(3\right)$

Put $x=1$ in eqn$\left(3\right)$ we get

${(1+1)}^n+n\times 1{(1+1)}^{n-1}=c_0+2c_{1}x+3c_2{1}^2+4c_3{1}^3+...+(n+1)c_n{1}^n$

$\Rightarrow 2^n+n.2^{n-1}=c_0+2c_1+3c_2+4c_3+...+(n+1)c_n$

$\Rightarrow c_0+2c_1+3c_2+4c_3+...+(n+1)c_n=2^{n-1}(n+2)$ is proved