Tìm M = 1 + 1/2 + 1/2^2 + 1/2^3 + . . . + 1/2^99 + 1/2^100

10/27

Tìm \[M = 1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{99}}}} + \frac{1}{{{2^{100}}}}\]

\[\frac{1}{{{2^{99}}}}\]

\[\frac{{{2^{101}} - 2}}{{{2^{100}}}}\]

\[\frac{{{2^{101}} + 1}}{{{2^{100}}}}\]

\[\frac{{{2^{101}} - 1}}{{{2^{100}}}}\]

Giải thích

\[\begin{array}{*{20}{l}}{M = 1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{99}}}} + \frac{1}{{{2^{100}}}}}\\{2M = 2.\left( {1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{99}}}} + \frac{1}{{{2^{100}}}}} \right)}\\{ = 2 + 1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{99}}}}}\end{array}\]

Ta có:

\[M = 2M - M\]

\[ = \left( {2 + 1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{98}}}} + \frac{1}{{{2^{99}}}}} \right) - \left( {1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{99}}}} + \frac{1}{{{2^{100}}}}} \right)\]

\[ = 2 - \frac{1}{{{2^{100}}}}\]

\[ = \frac{{{2^{101}} - 1}}{{{2^{100}}}}\]

Đáp án cần chọn là: D