Evaluate a sum with three different structures

$\begin{aligned}&S\\ & =\sum_{r=2}^n2^{-r}+2n\sum_{r=2}^nr\\ &{}+n^2\sum_{r=2}^n1.\end{aligned}$

Each has n−1 terms.

The geometric part has first term $1/4$, ratio $1/2$:

$\begin{aligned}&\sum_{r=2}^n2^{-r}\\ & =\frac{(1/4)[1-(1/2)^{n-1}]}{1-1/2}\\ & =\frac12-2^{-n}.\end{aligned}$

Also $\sum_{r=2}^nr=(n-1)(n+2)/2$ and $\sum_{r=2}^n1=n-1$.

$\begin{aligned}&S\\ & =\frac12-2^{-n}+n(n-1)(n+2)\\ &{}+n^2(n-1)\\ & =\frac12-2^{-n}\\ &{}+2n(n-1)(n+1)\\ & =\frac12-2^{-n}+2n(n^2-1).\end{aligned}$

For the printed sum assume integer $n\ge2$.