Noticing that the denominator of each term forms a geometric sequence and the numerator forms an arithmetic sequence. Tackling such a sequence is one basic sequence discussed in the book %%HREF%%Power Calculation%%Home/35-books/97-book-power-calculation%%. Multiplying both sides by $1/2$: $$\frac{1}{2}\times S_n = \frac{2}{2^2}+\frac{3}{2^3}+\cdots + \frac{n}{2^n}+\frac{n+1}{2^{n+1}}$$ Subtracting this from the original relation yields: \begin{align*} \frac{1}{2}\times S_n &= \frac{2}{2}+\Big(\frac{1}{2^2}+\frac{1}{2^3}+\cdots +\frac{1}{2^n}\Big)-\frac{n+1}{2^{n+1}}\\ &=1+\frac{1}{2^2}\times\frac{1-1/2^{n-1}}{1-1/2}-\frac{n+1}{2^{n+1}}\\ &=\frac{3}{2}-\frac{n+3}{2^{n+1}}\\ \therefore\quad S_n &=\boxed{3-\frac{n+3}{2^n}} \end{align*}