Bài 6 trang 83 Tài liệu dạy – học toán 6 tập 2

Đề bài

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) \({1 \over 3} – {2 \over {{3^2}}} + {3 \over {{3^3}}} – {4 \over {{3^4}}} + … + {{99} \over {{3^{99}}}} – {{100} \over {{3^{100}}}} < {3 \over {16}}\).

Lời giải chi tiết

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:

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: \({{16} \over 9}A < {1 \over 3} \Rightarrow A < {1 \over 3}:{{16} \over 9} = {3 \over {16}}.\)

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