(0,5 điểm) Cho biểu thức A = − 1 3 + 1 3 2 − 1 3 3 + 1 3 4 − 1 3 5 + . . . + 1 3 100 . Tính giá trị của biểu thức B = 4 | A | + 1 3 100 .

Hướng dẫn giải

Ta có: \(A = - \frac{1}{3} + \frac{1}{{{3^2}}} - \frac{1}{{{3^3}}} + \frac{1}{{{3^4}}} - \frac{1}{{{3^5}}} + ... + \frac{1}{{{3^{100}}}}\)

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

Suy ra \(3A + A = - 1 + \frac{1}{3} - \frac{1}{{{3^2}}} + \frac{1}{{{3^3}}} - \frac{1}{{{3^4}}} + ... + \frac{1}{{{3^{99}}}} + \left( { - \frac{1}{3} + \frac{1}{{{3^2}}} - \frac{1}{{{3^3}}} + \frac{1}{{{3^4}}} - \frac{1}{{{3^5}}} + ... + \frac{1}{{{3^{100}}}}} \right)\)

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

\(4A = - 1 + \frac{1}{{{3^{100}}}}\)

Suy ra \(A = \frac{1}{4}\left( { - 1 + \frac{1}{{{3^{100}}}}} \right)\).

Do \({3^{100}} > 1\) suy ra \(\frac{1}{{{3^{100}}}} < 1\) nên \(\frac{1}{{{3^{100}}}} - 1 < 0\) suy ra \(\frac{1}{4}\left( { - 1 + \frac{1}{{{3^{100}}}}} \right) < 0\) hay \(A < 0\).

Suy ra \(\left| A \right| = - A\).

Do đó, \(B = 4\left| A \right| + \frac{1}{{{3^{100}}}} = - 4A + \frac{1}{{{3^{100}}}} = - 4.\frac{1}{4}\left( {\frac{1}{{{3^{100}}}} - 1} \right) + \frac{1}{{{3^{100}}}}\)

\( = - \left( {\frac{1}{{{3^{100}}}} - 1} \right) + \frac{1}{{{3^{100}}}} = - \frac{1}{{{3^{100}}}} + 1 + \frac{1}{{{3^{100}}}} = 1 + \left( { - \frac{1}{{{3^{100}}}} + \frac{1}{{{3^{100}}}}} \right) = 1\).

Vậy \(B = 1\).