$(\dfrac{1}{5} + \dfrac{1}{15} + \dfrac{1}{25} + ... + \dfrac{1}{1985}) < \dfrac{9}{20}$
Edited by Pirates, 15-04-2010 - 14:31.
T7 (:D)
Started By rainy_o0o_sunny1, 12-04-2010 - 21:18
#1
Posted 12-04-2010 - 21:18
rainy_o0o_sunny1
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CMR:
$(\dfrac{1}{5} + \dfrac{1}{15} + \dfrac{1}{25} + ... + \dfrac{1}{1985}) < \dfrac{9}{20}$
$(\dfrac{1}{5} + \dfrac{1}{15} + \dfrac{1}{25} + ... + \dfrac{1}{1985}) < \dfrac{9}{20}$
#2
Posted 23-04-2010 - 18:19
hiep ga
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oh bài này làm rùi nhưng ko nhớ chờ ra hỏi cái đãCMR:
$(\dfrac{1}{5} + \dfrac{1}{15} + \dfrac{1}{25} + ... + \dfrac{1}{1985}) < \dfrac{9}{20}$
Poof
#3
Posted 23-04-2010 - 20:29
pth_tdn
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Cách em hơi dở >"<...
$A=\dfrac{1}{5}(1+\dfrac{1}{3}+...)$
$=\dfrac{1}{5}[1+\dfrac{1}{3}+\dfrac{1}{5}+(\dfrac{1}{7}+...+\dfrac{1}{15})+(\dfrac{1}{17}+...+\dfrac{1}{25})+(\dfrac{1}{27}+...+\dfrac{1}{35})+(\dfrac{1}{37}+...+\dfrac{1}{45})+...]$
$>\dfrac{1}{5}(1+\dfrac{1}{3}+\dfrac{1}{5}+5.\dfrac{1}{15}+5.\dfrac{1}{25}+5.\dfrac{1}{35}+5.\dfrac{1}{45})$
$>\dfrac{1}{5}.(1+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{1}{7}+\dfrac{1}{9})=\dfrac{1}{5}.(\dfrac{31}{15}+\dfrac{16}{63})>\dfrac{1}{5}.(2+\dfrac{1}{4})=\dfrac{9}{20}$
$A=\dfrac{1}{5}(1+\dfrac{1}{3}+...)$
$=\dfrac{1}{5}[1+\dfrac{1}{3}+\dfrac{1}{5}+(\dfrac{1}{7}+...+\dfrac{1}{15})+(\dfrac{1}{17}+...+\dfrac{1}{25})+(\dfrac{1}{27}+...+\dfrac{1}{35})+(\dfrac{1}{37}+...+\dfrac{1}{45})+...]$
$>\dfrac{1}{5}(1+\dfrac{1}{3}+\dfrac{1}{5}+5.\dfrac{1}{15}+5.\dfrac{1}{25}+5.\dfrac{1}{35}+5.\dfrac{1}{45})$
$>\dfrac{1}{5}.(1+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{1}{7}+\dfrac{1}{9})=\dfrac{1}{5}.(\dfrac{31}{15}+\dfrac{16}{63})>\dfrac{1}{5}.(2+\dfrac{1}{4})=\dfrac{9}{20}$
Edited by pth_tdn, 23-04-2010 - 20:31.
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