Câu hỏi:

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So sánh A=12122+123124+...+129912100A = \frac{1}{2} - \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} - \frac{1}{{{2^4}}} + ... + \frac{1}{{{2^{99}}}} - \frac{1}{{{2^{100}}}}13\frac{1}{3}.

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Lời giải

Hướng dẫn giải:

Ta có:

A=12122+123124+...+129912100A = \frac{1}{2} - \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} - \frac{1}{{{2^4}}} + ... + \frac{1}{{{2^{99}}}} - \frac{1}{{{2^{100}}}}

Suy ra 2A=2(12122+123124+...+129912100)2A = 2\left( {\frac{1}{2} - \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} - \frac{1}{{{2^4}}} + ... + \frac{1}{{{2^{99}}}} - \frac{1}{{{2^{100}}}}} \right)

2A=112+122123+...+129812992A = 1 - \frac{1}{2} + \frac{1}{{{2^2}}} - \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{98}}}} - \frac{1}{{{2^{99}}}}

2A=112+122123+...+129812992A = 1 - \frac{1}{2} + \frac{1}{{{2^2}}} - \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{98}}}} - \frac{1}{{{2^{99}}}}

Do đó 2A+A=(112+122123+...+12981299)+(12122+123124+...+129912100)2A + A = \left( {1 - \frac{1}{2} + \frac{1}{{{2^2}}} - \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{98}}}} - \frac{1}{{{2^{99}}}}} \right) + \left( {\frac{1}{2} - \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} - \frac{1}{{{2^4}}} + ... + \frac{1}{{{2^{99}}}} - \frac{1}{{{2^{100}}}}} \right)

Suy ra 3A=1121003A = 1 - \frac{1}{{{2^{100}}}}

A=1313.2100A = \frac{1}{3} - \frac{1}{{{{3.2}^{100}}}}

13.2100>0\frac{1}{{{{3.2}^{100}}}} > 0 nên 1313.2100<13\frac{1}{3} - \frac{1}{{{{3.2}^{100}}}} < \frac{1}{3}.

Vậy A<13A < \frac{1}{3}.

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