Cho S = 1/2+ 1/2^2 + 1/2^3 + ... + 1/2^2022. So sánh S với 1.

10/10

Cho \[S = \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{2022}}}}\]. So sánh \[S\] với 1.

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Giải thích

Ta có: \[S = \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{2022}}}}\].

Suy ra \[2S = 2\,\,.\,\,\left( {\frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{2022}}}}} \right)\]

\[ = \frac{2}{2} + \frac{2}{{{2^2}}} + \frac{2}{{{2^3}}} + ... + \frac{2}{{{2^{2022}}}}\]\[ = 1 + \frac{1}{2} + \frac{1}{{{2^2}}} + ... + \frac{1}{{{2^{2021}}}}\].

Ta có \[S = \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{2022}}}}\] và \[2S = 1 + \frac{1}{2} + \frac{1}{{{2^2}}} + ... + \frac{1}{{{2^{2021}}}}\].

Suy ra \(2S - S = \left( {1 + \frac{1}{2} + \frac{1}{{{2^2}}} + ... + \frac{1}{{{2^{2021}}}}} \right) - \left( {\frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{2022}}}}} \right)\) .

Hay \(S = 1 + \frac{1}{2} + \frac{1}{{{2^2}}} + ... + \frac{1}{{{2^{2021}}}} - \frac{1}{2} - \frac{1}{{{2^2}}} - \frac{1}{{{2^3}}} - ... - \frac{1}{{{2^{2022}}}}\)

\( = 1 + \left( {\frac{1}{2} - \frac{1}{2}} \right) + \left( {\frac{1}{{{2^2}}} - \frac{1}{{{2^2}}}} \right) + \left( {\frac{1}{{{2^3}}} - \frac{1}{{{2^3}}}} \right) + ... + \left( {\frac{1}{{{2^{2021}}}} - \frac{1}{{{2^{2021}}}}} \right) - \frac{1}{{{2^{2022}}}}\)

\( = 1 - \frac{1}{{{2^{2022}}}} = \frac{{{2^{2022}} - 1}}{{{2^{2022}}}}\).

Mà \[{2^{2022}}--1 < {2^{2022}}\] nên \[\frac{{{2^{2022}} - 1}}{{{2^{2022}}}} < 1\];

Vậy \[S = \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{2022}}}} < 1\].