Chứng minh P = 1/3^1 + 2/3^2 + 3/3^3 + + 100/3^100 < 3/4
Câu hỏi:
Chứng minh \(P = \frac{1}{{{3^1}}} + \frac{2}{{{3^2}}} + \frac{3}{{{3^3}}} + ... + \frac{{100}}{{{3^{100}}}} < \frac{3}{4}\).
Trả lời:
\(P = \frac{1}{{{3^1}}} + \frac{2}{{{3^2}}} + \frac{3}{{{3^3}}} + ... + \frac{{100}}{{{3^{100}}}}\)
⇒ \(\frac{1}{3}P = \frac{1}{{{3^2}}} + \frac{2}{{{3^3}}} + \frac{3}{{{3^4}}} + ... + \frac{{100}}{{{3^{101}}}}\)
\(P - \frac{1}{3}P = \frac{1}{{{3^1}}} + \left( {\frac{2}{{{3^2}}} - \frac{1}{{{3^2}}}} \right) + \left( {\frac{3}{{{3^3}}} - \frac{2}{{{3^3}}}} \right) + + ... + \left( {\frac{{100}}{{{3^{100}}}} - \frac{{99}}{{{3^{100}}}}} \right) - \frac{{100}}{{{3^{101}}}}\)
\(\frac{2}{3}P = \frac{1}{{{3^1}}} + \frac{1}{{{3^2}}} + \frac{1}{{{3^3}}} + ... + \frac{1}{{{3^{100}}}} - \frac{{100}}{{{3^{101}}}}\) (*)
Đặt \(S = \frac{1}{{{3^1}}} + \frac{1}{{{3^2}}} + \frac{1}{{{3^3}}} + ... + \frac{1}{{{3^{100}}}}\)
(*) trở thành: \(\frac{2}{3}P = S - \frac{{100}}{{{3^{101}}}}\)(1)
Xét \(S = \frac{1}{{{3^1}}} + \frac{1}{{{3^2}}} + \frac{1}{{{3^3}}} + ... + \frac{1}{{{3^{100}}}}\)
⇒\(\frac{1}{3}S = \frac{1}{{{3^2}}} + \frac{1}{{{3^3}}} + \frac{1}{{{3^4}}} + ... + \frac{1}{{{3^{101}}}}\)
⇒ \(S - \frac{1}{3}S = \frac{1}{{{3^1}}} - \frac{1}{{{3^{101}}}}\)
⇔ \(S - \frac{1}{3}S = \frac{1}{{{3^1}}} - \frac{1}{{{3^{101}}}}\)
⇔ \(\frac{2}{3}S = \frac{1}{{{3^1}}} - \frac{1}{{{3^{101}}}}\)
\(S = \frac{3}{2}.\left( {\frac{1}{{{3^1}}} - \frac{1}{{{3^{101}}}}} \right)\)(2)
Thay (2) vào (1) ta có:
\(\frac{2}{3}P = \frac{3}{2}.\left( {\frac{1}{{{3^1}}} - \frac{1}{{{3^{101}}}}} \right) - \frac{{100}}{{{3^{101}}}}\)
\(P = \frac{9}{4}.\left( {\frac{1}{{{3^1}}} - \frac{1}{{{3^{101}}}}} \right) - \frac{{100}}{{{3^{101}}}} = \frac{3}{4} - \frac{9}{{{{4.3}^{101}}}} - \frac{{100}}{{{3^{101}}}} < \frac{3}{4}\)
Vậy P < \(\frac{3}{4}\).
