Tính tổng biểu thức M = 1/ 5 + 1 /5 ^2 + 1/ 5^ 3 + . . . + 1/ 5 ^10

Ta có

\(\begin{array}{l}M = \frac{1}{{{5^{10}}}} + ... + \frac{1}{{{5^3}}} + \frac{1}{{{5^2}}} + \frac{1}{5} \Leftrightarrow M + 1 = {\left( {\frac{1}{5}} \right)^{10}} + ... + {\left( {\frac{1}{5}} \right)^3} + {\left( {\frac{1}{5}} \right)^2} + \frac{1}{5} + 1\\ \Leftrightarrow \left( {M + 1} \right)\left( {\frac{1}{5} - 1} \right) = \left( {\frac{1}{5} - 1} \right)\left[ {{{\left( {\frac{1}{5}} \right)}^{10}} + ... + {{\left( {\frac{1}{5}} \right)}^3} + {{\left( {\frac{1}{5}} \right)}^2} + \frac{1}{5} + 1} \right]\\ \Leftrightarrow - \frac{4}{5}\left( {M + 1} \right) = {\left( {\frac{1}{5}} \right)^{11}} - 1 \Leftrightarrow M + 1 = \frac{{{{5.5}^{10}} - 1}}{{{{4.5}^{10}}}} \Leftrightarrow M = \frac{{{5^{10}} - 1}}{{{{4.5}^{10}}}} = \frac{1}{4}\left[ {1 - {{\left( {\frac{1}{5}} \right)}^{10}}} \right]\end{array}\)