Sum 4

We note that:

$\begin{aligned} \sum_{r=0}^n {n \choose r}x^r & = (1+x)^n & \small \color{#3D99F6} \text{Differentiate both sides w.r.t. }x \\ \frac d{dx} \sum_{r=0}^n {n \choose r}x^r & = \frac d{dx} (1+x)^n \\ \sum_{\color{#D61F06}r=0}^n r{n \choose r}x^{r-1} & = n(1+x)^{n-1} & \small \color{#3D99F6} \text{Putting }x=1 \\ \sum_{\color{#3D99F6}r=1}^n r{n \choose r} & = \boxed{n2^{n-1}} \end{aligned}$

Actually here see..n $\dbinom {n}{1}$ + n $\dbinom {n}{2}$ ..... +n $\dbinom {n}{n}$

so n( $\dbinom {n}{1}$ + $\dbinom {n}{2}$ ..... + $\dbinom {n}{n}$ )

Now Expansion of $(1+1)^n$ = $2^n$ = ( $\dbinom{n}{0}$ + $\dbinom {n}{1}$ + $\dbinom {n}{2}$ ..... + $\dbinom {n}{n}$ )

So $2^n-1$ =( $\dbinom {n}{1}$ + $\dbinom {n}{2}$ ..... + $\dbinom {n}{n}$ )

Hence n( $\dbinom {n}{1}$ + $\dbinom {n}{2}$ ..... + $\dbinom {n}{n}$ ) = $n2^n-n$ = $\boxed{n(2^n-1)}$