Chứng minh biểu thức dưới đây: 1/3 - 2/3^2 + 3/3^3 - 4/3^4 + ... + 99/3^99 - 100/3^100 < 3/16

45/62

Chứng minh:

\[\frac{1}{3} - \frac{2}{{{3^2}}} + \frac{3}{{{3^3}}} - \frac{4}{{{3^4}}} + ... + \frac{{99}}{{{3^{99}}}} - \frac{{100}}{{{3^{100}}}} < \frac{3}{{16}}\].

Nhập câu trả lời ở đây

0/3000 ký tự

Giải thích

Đặt \[A = \frac{1}{3} - \frac{2}{{{3^2}}} + \frac{3}{{{3^3}}} - \frac{4}{{{3^4}}} + ... + \frac{{99}}{{{3^{99}}}} - \frac{{100}}{{{3^{100}}}}\]

Suy ra \[3A = 1 - \frac{2}{3} + \frac{3}{{{3^2}}} - \frac{4}{{{3^3}}} + ... + \frac{{99}}{{{3^{98}}}} - \frac{{100}}{{{3^{99}}}}\]

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

\[4A = 1 - \frac{2}{3} + \frac{3}{{{3^2}}} - \frac{4}{{{3^3}}} + ... + \frac{{99}}{{{3^{98}}}} - \frac{{100}}{{{3^{99}}}}\]

\[ = 1 - \frac{1}{3} + \frac{1}{{{3^2}}} - \frac{1}{{{3^3}}} + ... - \frac{1}{{{3^{99}}}} - \frac{{100}}{{{3^{100}}}}\]

Suy ra \[12A = 3 - 1 + \frac{1}{3} - \frac{1}{{{3^2}}} + ... - \frac{1}{{{3^{98}}}} - \frac{{100}}{{{3^{99}}}}\]

Nên \[4A + 12A = 3 - \frac{{100}}{{{3^{99}}}} - \frac{1}{{{3^{99}}}} - \frac{{100}}{{{3^{100}}}} < 3\]

Hay \[16A < 3\] suy ra \[A < \frac{3}{{16}}.\]

Vậy \[\frac{1}{3} - \frac{2}{{{3^2}}} + \frac{3}{{{3^3}}} - \frac{4}{{{3^4}}} + ... + \frac{{99}}{{{3^{99}}}} - \frac{{100}}{{{3^{100}}}} < \frac{3}{{16}}.\]