If ∑(r = 0 to n) (nCr)^-1 = s,
then prove that ∑(r = 0 to n) (2r + 1)/(nCr) = (n+1)s
nCr is n choose r.
Let $\displaystyle \sum_{r=0}^n \frac{1}{{n\choose r}} = s$ and $\displaystyle A = \sum_{r=0}^n \frac{r}{{n\choose r}}$.
Notice that, $\displaystyle \sum_{r=0}^n \frac{r}{{n\choose r}} = \sum_{r=0}^n \frac{r}{{n\choose {n-r}}}$ now let $\displaystyle t=n-r$ and so $\displaystyle \sum_{r=0}^n \frac{r}{{n\choose r}} = \sum_{r=0}^n \frac{n-t}{{n\choose t}}$.
Therefore, $\displaystyle 2\sum_{r=0}^{n} \frac{r}{{n\choose r}} = \sum_{t=0}^n \frac{n}{{n\choose r}} = ns \implies A = \tfrac{1}{2}ns$
Thus, $\displaystyle \sum_{r=0}^n \frac{2r+1}{{n\choose r}} = 2\sum_{r=0}^n \frac{r}{{n\choose r}} + \sum_{r=0}^n \frac{1}{{n\choose r}} = 2\left( \tfrac{1}{2}ns \right) + s = ns + s = s(n+1)$.