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Chứng minh rằng :
a) \({1 \over 2} – {1 \over 4} + {1 \over 8} – {1 \over {16}} + {1 \over {32}} – {1 \over {64}} < {1 \over 3}\).
b) \({1 \over 3} – {2 \over {{3^2}}} + {3 \over {{3^3}}} – {4 \over {{3^4}}} + … + {{99} \over {{3^{99}}}} – {{100} \over {{3^{100}}}} < {3 \over {16}}\).
a)Cách 1:
Đặt \(A = {1 \over 2} – {1 \over 4} + {1 \over 8} – {1 \over {16}} + {1 \over {32}} – {1 \over {64}} \Rightarrow 2A = 1 – {1 \over 2} + {1 \over 4} – {1 \over 8} + {1 \over {16}} – {1 \over {32}}\)
\(\eqalign{ & 2A + A = \left( {1 – {1 \over 2} + {1 \over 4} – {1 \over 8} + {1 \over {16}} – {1 \over {32}}} \right) + \left( {{1 \over 2} – {1 \over 4} + {1 \over 8} – {1 \over {16}} + {1 \over {32}} – {1 \over {64}}} \right) \cr & 3A = 1 – {1 \over 2} + {1 \over 2} + {1 \over 4} – {1 \over 4} – {1 \over 8} + {1 \over 8} + {1 \over {16}} – {1 \over {16}} – {1 \over {32}} + {1 \over {32}} – {1 \over {64}} \cr & 3A = 1 – {1 \over {64}} \Leftrightarrow 3A = {{63} \over {64}}. \cr} \)
Mà \({{63} \over {64}} < 1.\) Nên 3A < 1. Vậy \(A < {1 \over 3}.\)
Cách 2:
\({1 \over 2} – {1 \over 4} + {1 \over 8} – {1 \over {16}} + {1 \over {32}} – {1 \over {64}} = {{32 – 16 + 8 – 4 + 2 – 1} \over {64}} = {{21} \over {64}} < {{21} \over {63}} = {1 \over 3}.\)
b) Cách 1:
Advertisements (Quảng cáo)
Đặt \(A = {1 \over 3} – {2 \over {{3^2}}} + {3 \over {{3^3}}} – {4 \over {{3^4}}} + … + {{99} \over {{3^{99}}}} – {{100} \over {{3^{100}}}} \Rightarrow {1 \over 3}A = {1 \over {{3^2}}} – {2 \over {{3^3}}} + {3 \over {{3^4}}} – {4 \over {{3^5}}} + … + {{99} \over {{3^{100}}}} – {{100} \over {{3^{101}}}}\)
Do đó: \(A + {1 \over 3}A = {1 \over 3} – {1 \over {{3^2}}} + {1 \over {{3^3}}} – {1 \over {{3^4}}} + … {1 \over {{3^{100}}}} – {{100} \over {{3^{101}}}}\)
\(4A = 2 – {1 \over 3} + {1 \over {{3^2}}} – {1 \over {{3^3}}} + … {1 \over {{3^{99}}}} – {{100} \over {{3^{100}}}} \Rightarrow 12A = 3 – 1 + {1 \over 3} – {1 \over {{3^2}}} + … {1 \over {{3^{98}}}} – {{100} \over {{3^{99}}}}\)
Do đó: \(16A = 3 – {{101} \over {{3^{99}}}} – {{100} \over {{3^{100}}}}.\) Mà \(3 – {{101} \over {{3^{99}}}} – {{100} \over {{3^{100}}}} < 3.\) Nên 16A < 3.
Vậy \(A < 3.{1 \over {16}} = {3 \over {16}}.\)
Cách 2:
Đặt \(A = {1 \over 3} – {2 \over {{3^2}}} + {3 \over {{3^3}}} – {4 \over {{3^4}}} + … + {{99} \over {{3^{99}}}} – {{100} \over {{3^{100}}}} \Rightarrow {2 \over 3}A = + {2 \over {{3^2}}} – {4 \over {{3^2}}} + {6 \over {{3^4}}} – … {{196} \over {{3^{99}}}} + {{198} \over {{3^{100}}}} – {{200} \over {{3^{101}}}}\)
\({1 \over {{3^2}}}A = + {1 \over {{3^3}}} – {2 \over {{3^4}}} + … + {{97} \over {{3^{99}}}} – {{98} \over {{3^{100}}}} + {{99} \over {{3^{101}}}} – {{100} \over {{3^{102}}}} – {{101} \over {{3^{101}}}} – {{100} \over {{3^{102}}}} \Leftrightarrow {{16} \over 9}A = {1 \over 3}\)
Ta có: \({1 \over 3} – {{101} \over {{3^{101}}}} – {{100} \over {{3^{102}}}} < {1 \over 3}.\) Do đó: \({{16} \over 9}A < {1 \over 3} \Rightarrow A < {1 \over 3}:{{16} \over 9} = {3 \over {16}}.\)