Tính tổng: . S=1/5+1/5^2+1/5^3+...+1/5^100

Ta có:

\[5S = 1 + \frac{1}{5} + \frac{1}{{{5^2}}} + \frac{1}{{{5^3}}} + ... + \frac{1}{{{5^{99}}}}\]

\[5S - S = \left( {1 + \frac{1}{5} + \frac{1}{{{5^2}}} + \frac{1}{{{5^3}}} + ... + \frac{1}{{{5^{99}}}}} \right) - \left( {\frac{1}{5} + \frac{1}{{{5^2}}} + \frac{1}{{{5^3}}} + ... + \frac{1}{{{5^{100}}}}} \right)\]

\[4S = 1 - \frac{1}{{{5^{100}}}} = \frac{{{5^{100}} - 1}}{{{5^{100}}}}\]

Suy ra \[S = \frac{{{5^{100}} - 1}}{{4 \cdot {5^{100}}}}.\]