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: \[{{16} \over 9}A < {1 \over 3} \Rightarrow A < {1 \over 3}:{{16} \over 9} = {3 \over {16}}.\]