S_n = \sum_{k=1}^n \frac{2^{k+1}}{k+1} {n\choose k} = \frac{1}{n+1} \sum_{k=1}^n \frac{2^{k+1} (n+1)n!}{(k+1)k!(n-k)!}  =   \frac{1}{n+1} \sum_{k=1}^n 2^{k+1} {n+1\choose k+1} = \frac{1}{n+1}\left ( \sum_{k=0}^n 2^{k+1} {n+1\choose k+1} - 2(n+1)\right) = \frac{1}{n+1} \biggl(3^{n+1}-1 -2n -2\biggr) = \frac{3^{n+1} -2n -3}{n+1}