(0,5 điểm) Cho A = 1 3 − 2 3 2 + 3 3 3 − 4 3 4 + . . . + 99 3 99 − 100 3 100 . Chứng minh A < 3 16 .

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(0,5 điểm) Cho \(A = \frac{1}{3} - \frac{2}{{{3^2}}} + \frac{3}{{{3^3}}} - \frac{4}{{{3^4}}} + ... + \frac{{99}}{{{3^{99}}}} - \frac{{100}}{{{3^{100}}}}\). Chứng minh \(A < \frac{3}{{16}}.\)

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Hướng dẫn giải

Ta có: \(A = \frac{1}{3} - \frac{2}{{{3^2}}} + \frac{3}{{{3^3}}} - \frac{4}{{{3^4}}} + ... + \frac{{99}}{{{3^{99}}}} - \frac{{100}}{{{3^{100}}}}\)

\(3A = \left( {1 - \frac{2}{3} + \frac{3}{{{3^2}}} - \frac{4}{{{3^3}}} + ... + \frac{{99}}{{{3^{98}}}} - \frac{{100}}{{{3^{99}}}}} \right)\)

\(3A + A = \left( {1 - \frac{2}{3} + \frac{3}{{{3^2}}} - \frac{4}{{{3^3}}} + ... + \frac{{99}}{{{3^{98}}}} - \frac{{100}}{{{3^{99}}}}} \right) + \left( {\frac{1}{3} - \frac{2}{{{3^2}}} + \frac{3}{{{3^3}}} - \frac{4}{{{3^4}}} + ... + \frac{{99}}{{{3^{99}}}} - \frac{{100}}{{{3^{100}}}}} \right)\)

\(3A + A = 1 - \frac{1}{3} + \frac{1}{{{3^2}}} - \frac{1}{{{3^3}}} + ... + \frac{1}{{{3^{98}}}} - \frac{1}{{{3^{99}}}} - \frac{{100}}{{{3^{100}}}}\)

\(4A = \left( {1 - \frac{1}{3} + \frac{1}{{{3^2}}} - \frac{1}{{{3^3}}} + ... + \frac{1}{{{3^{98}}}} - \frac{1}{{{3^{99}}}}} \right) - \frac{{100}}{{{3^{100}}}}\)

Đặt \(B = 1 - \frac{1}{3} + \frac{1}{{{3^2}}} - \frac{1}{{{3^3}}} + ... + \frac{1}{{{3^{98}}}} - \frac{1}{{{3^{99}}}}\)

\(3B = 3 - 1 + \frac{1}{3} - \frac{1}{{{3^2}}} + \frac{1}{{{3^3}}} - \frac{1}{{{3^4}}} + ... + \frac{1}{{{3^{97}}}} - \frac{1}{{{3^{98}}}}\)

\(3B + B = 3 - \frac{1}{{{3^{99}}}}\)

\(4B = 3 - \frac{1}{{{3^{99}}}} < 3\) suy ra \(B < \frac{3}{4}.\)

Từ đây, suy ra \(4A = B - \frac{{100}}{{{3^{100}}}}\) nên \(4A < \frac{3}{4}\), do đó \(A < \frac{3}{{16}}\) (đpcm).