Sum of the first $$n$$ terms of the series $$\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{{15}}{{16}} + ......$$      is equal to

Solution :
$$\eqalign{ & {\text{Let }}S = \frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{{15}}{{16}} + ......n{\text{ terms}} \cr & {\text{ = }}\left( {1 - \frac{1}{2}} \right) + \left( {1 - \frac{1}{4}} \right) + \left( {1 - \frac{1}{8}} \right) + \left( {1 - \frac{1}{{16}}} \right).....n\,{\text{terms}} \cr & {\text{ = }}\left( {1 + 1 + 1 + .....n{\text{ terms}}} \right) - \left( {\frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + \frac{1}{{{2^4}}} + .... + \frac{1}{{{2^n}}}} \right) \cr & = n - \left[ {\frac{{\frac{1}{2}\left( {1 - \frac{1}{{{2^n}}}} \right)}}{{1 - \frac{1}{2}}}} \right] \cr & = n - 1 + {2^{ - n}} \cr} $$