1 + \(\frac{3}{2}+\frac{5}{4}+\frac{7}{8}\) + ..... is an arithmetico-geometric series with corresponding A.P. and G.P as: 

A.P. : 1 + 3 + 5 + 7 + ......

G.P. : 1 + \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\) + ...........

nth term of A.P. = (1 + (n – 1) 2) = (2n – 1)

nth term of G.P. = 1. \(\big(\frac{1}{2}\big)^{n-1}\) = \(\frac{1}{2^{n-1}}\)

∴ nth term of the given A.G.P. = \(\frac{2n-1}{2^{n-1}}\)

∴ S_n = 1 + \(\frac{3}{2}+\frac{5}{4}+\frac{7}{8}\) + ........ + \(\frac{2n-1}{2^{n-1}}\)

⇒ \(\frac{1}{2}\)S_n = \(\frac{1}{2}+\frac{3}{4}+\frac{5}{8}\) + ......+........+ \(\frac{2n-3}{2^{n-1}}\) + \(\frac{2n-1}{2^n}\)

On subtraction, we get

\(\big(1-\frac{1}{2}\big)\)S_n = 1 + \(\frac{2}{2}+\frac{2}{4}+\frac{2}{8}\) + ........... + \(\frac{2}{2^{n-1}}\) - \(\frac{2n-1}{2^n}\)

⇒ \(\frac{1}{2}\)S_n = 1 + 2 \(\bigg\{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\)+ ........+ \(\frac{1}{2^{n-1}}\bigg\}\) - \(\frac{2n-1}{2^n}\)

⇒ \(\frac{1}{2}\)S_n = 1 + 2 \(\bigg\{\frac{\big(\frac{1}{2}\big(1-\big(\frac{1}{2}\big)^n\big)}{1-\frac{1}{2}}\bigg\}\) - \(\frac{2n-1}{2^n}\) = 1 + 2 = \(\frac{4}{2^n}\) - \(\frac{2n-1}{2^n}\) = 3 - \(\frac{2n-3}{2^{n}}\)

∴ S_n = 6 - \(\frac{2n-3}{2^{n-1}}\).