Tính M = 1 + 1/2 + 1/2^2 + 1/2^3 + ... + 1/2^99 + 1/2^100 A. 1/2^99 B. 2^101 - 2/2^100 C. 2^101 + 1/2^100 D. 2^101 - 1/2^100
Giải thích
Trả lời:
\[M = 1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{99}}}} + \frac{1}{{{2^{100}}}}\]
\[2M = 2.\left( {1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{99}}}} + \frac{1}{{{2^{100}}}}} \right)\]
\[M = 2 + 1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{99}}}}\]
Ta có:
M = 2M – M
\[ = \left( {2 + 1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{99}}}}} \right) - \left( {1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{99}}}} + \frac{1}{{{2^{100}}}}} \right)\]
\[ = 2 - \frac{1}{{{2^{100}}}}\]
\[ = \frac{{{2^{101}} - 1}}{{{2^{100}}}}\]
Đáp án cần chọn là: D