(0,5 điểm) Rút gọn biểu thức: A = 1/3 + 2/3^2 + 3/3^3 + . . . + 99/3^99 + 100/3^100 .

Hướng dẫn giải

Ta có: \[3A = 1 + \frac{2}{3} + \frac{3}{{{3^2}}} + ... + \frac{{99}}{{{3^{98}}}} + \frac{{100}}{{{3^{99}}}}\]

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

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

Đặt \[B = 1 + \frac{1}{3} + \frac{1}{{{3^2}}} + ... + \frac{1}{{{3^{98}}}} + \frac{1}{{{3^{99}}}}\].

Ta có \[3B = 3 + 1 + \frac{1}{3} + ... + \frac{1}{{{3^{97}}}} + \frac{1}{{{3^{98}}}}\]

Suy ra \[3B - B = \left( {3 + 1 + \frac{1}{3} + ... + \frac{1}{{{3^{97}}}} + \frac{1}{{{3^{98}}}}} \right) - \left( {1 + \frac{1}{3} + \frac{1}{{{3^2}}} + ... + \frac{1}{{{3^{98}}}} + \frac{1}{{{3^{99}}}}} \right)\]

\[2B = 3 - \frac{1}{{{3^{99}}}}\]

\[B = \frac{3}{2} - \frac{1}{{{3^{99}} \cdot 2}}\]

Do đó \[2A = \frac{3}{2} - \frac{1}{{{3^{99}} \cdot 2}} - \frac{{100}}{{{3^{100}}}}\]

Khi đó, \[A = \frac{3}{4} - \frac{1}{{{3^{99}} \cdot 4}} - \frac{{100}}{{{3^{100}} \cdot 2}} = \frac{{3 \cdot {3^{100}} - 3 - 2 \cdot 100}}{{4 \cdot {3^{100}}}} = \frac{{3 \cdot {3^{100}} - 203}}{{4 \cdot {3^{100}}}}.\]