If $(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+\dots +C_n{x}^n,$ the value of $C_0+2C_1+3C_2+\dots +(n+1)C_n$ is 

Let $E=C_0+2C_1+3C_2+\dots +n{C}_{n-1}+(n+1)C_n \left(1\right)$
Using $C_r=C_{n-r}$, we can rewrite $\left(1\right)$ as
$E=(n+1)C_0+n{C}_1+(n-1)C_2+\dots +2C_{n-1}+C_n \left(2\right)$
Adding $\left(1\right)$ and $\left(2\right)$, we get

              $\begin{array}{l} 2E=(n+2)C_0+(n+2)C_1+(n+2)C_2+\dots \\ +(n+2)C_n\\ =(n+2)\left\{C_0+C_1+\dots +C_n\right\}=(n+2)2^n\\ \Rightarrow E=(n+2)2^{n-1}\end{array}$

Alternative Solution

We have $C_{0}x+C_1{x}^2+C_2{x}^3+\dots +C_n{x}^{n+1}=x(1+x{)}^n$
Differentiating both the sides, we get

             $\begin{array}{l}{C}_0+2C_{1}x+3C_2{x}^2+\dots +(n+1)C_n{x}^n\\ =(1+x{)}^n+nx(1+x{)}^{n-1}\end{array}$      (1)

Putting $x=1,$ we get
          $\begin{array}{c}{C}_0+2C_1+3C_2+\dots +(n+1)C_n\\ =2^n+n\left(1\right)2^{n-1}=(n+2)2^{n-1}\end{array}$