(0,5 điểm) Tính tổng: \(S = \left( {1 - \frac{1}{2}} \right) + \left( {1 - \frac{1}{4}} \right) + \left( {1 - \frac{1}{8}} \right) + \cdots + \left( {1 - \frac{1}{{{2^n}}}} \right).\) Tính \({S_{10}}.\)

Ta có \(S = \left( {1 - \frac{1}{2}} \right) + \left( {1 - \frac{1}{4}} \right) + \left( {1 - \frac{1}{8}} \right) + \cdots + \left( {1 - \frac{1}{{{2^{10}}}}} \right)\)

\( = 10 - \left( {\frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + \cdots + \frac{1}{{{2^{10}}}}} \right)\).

Đặt \(M = \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + \cdots + \frac{1}{{{2^{10}}}}\).

Ta có \(2M = 1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + \cdots + \frac{1}{{{2^9}}}\).

Khi đó \(2M - M = M = \left( {1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + \cdots + \frac{1}{{{2^9}}}} \right) - \left( {\frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + \cdots + \frac{1}{{{2^{10}}}}} \right) = 1 - \frac{1}{{{2^{10}}}}\).

Do đó \[S = 10 - 1 + \frac{1}{{{2^{10}}}} = 9 + \frac{1}{{{2^{10}}}}.\]