Đề bài

So sánh S với 2, biết \(S = \frac{1}{2} + \frac{2}{{{2^2}}} + \frac{3}{{{2^3}}} +  \ldots  + \frac{{2023}}{{{2^{2023}}}}\).

Phương pháp giải

Nhân hai vế của S với 2 để rút gọn S.

Lời giải của GV Loigiaihay.com

\(S = \frac{1}{2} + \frac{2}{{{2^2}}} + \frac{3}{{{2^3}}} +  \ldots  + \frac{{2023}}{{{2^{2023}}}}\)

\(2S = 1 + \frac{2}{2} + \frac{3}{{{2^2}}} + \frac{4}{{{2^3}}} +  \ldots  + \frac{{2023}}{{{2^{2022}}}}\)

\(2S - S = \left(1 + \frac{2}{2} + \frac{3}{{{2^2}}} + \frac{4}{{{2^3}}} +  \ldots  + \frac{{2023}}{{{2^{2022}}}}\right) - \left(\frac{1}{2} + \frac{2}{{{2^2}}} + \frac{3}{{{2^3}}} +  \ldots  + \frac{{2023}}{{{2^{2023}}}}\right)\)

\(2S - S = 1 + \left(\frac{2}{2} - \frac{1}{2}\right) + \left(\frac{3}{{{2^2}}} - \frac{2}{{{2^2}}}\right) + \left(\frac{4}{{{2^3}}} - \frac{3}{{{2^3}}}\right) +  \ldots  + \left(\frac{{2023}}{{{2^{2022}}}} - \frac{{2022}}{{{2^{2022}}}}\right) - \frac{{2023}}{{{2^{2023}}}}\)

\(2S - S = 1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} +  \ldots  + \frac{1}{{{2^{2022}}}} - \frac{{2023}}{{{2^{2023}}}}\)

\(S = 1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} +  \ldots  + \frac{1}{{{2^{2022}}}} - \frac{{2023}}{{{2^{2023}}}}\)

\(2S = 2 + 1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} +  \ldots  + \frac{1}{{{2^{2021}}}} - \frac{{2023}}{{{2^{2022}}}}\)

\(2S - S = \left(2 + 1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} +  \ldots  + \frac{1}{{{2^{2021}}}} - \frac{{2023}}{{{2^{2022}}}}\right) - \left(1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} +  \ldots  + \frac{1}{{{2^{2022}}}} - \frac{{2023}}{{{2^{2023}}}}\right)\)

\(2S - S = 2 + \left(1 - 1\right) + \left(\frac{1}{2} - \frac{1}{2}\right) + \left(\frac{1}{{{2^2}}} - \frac{1}{{{2^2}}}\right) +  \ldots  + \left(- \frac{{2023}}{{{2^{2022}}}} -\frac{1}{{{2^{2022}}}}\right) - \frac{{2023}}{{{2^{2023}}}}\)

\(2S - S = 2 - \frac{{2024}}{{{2^{2022}}}} + \frac{{2023}}{{{2^{2023}}}}\)

\(S = 2 - \frac{{4048 - 2023}}{{{2^{2023}}}}\)

Vậy \(S < 2\).