Re: [中學] 一題數列
※ 引述《sendohandy (腦殘眼殘的老師)》之銘言:
: p=1/52*102 + 1/53*101 + ... +1/102*52
: q=1/1*2 + 1/3*4 + ... + 1/101*102
: 求q/p=
: 手機排版請見諒,謝謝
q = (1/1 - 1/2) + (1/3 - 1/4) + ... + (1/101 -1/102)
= (1/1 + 1/2 + 1/3 + ... + 1/101 + 1/102) - 2(1/2 + 1/4 + 1/6 + ... +1/102)
= (1/1 + 1/2 + 1/3 + ... + 1/101 + 1/102) - (1/1 + 1/2 + 1/3 + ... +1/51)
= (1/52 + 1/53 + 1/54 + ... + 1/101 + 1/102)
= (52+102)/(52*102) + (53+101)/(53*101) + ... + (76+78)/(76*78) + 1/77
= 154( 1/(52*102) + 1//(53*101) + ... + 1/(76*78) + 1/(77*77*2) )
= 77(1/(52*102) + 1/(53*101) + ... +1/(102*52))
= 77p
q/p = 77
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04/16 19:56, , 1F
04/16 19:56, 1F
原題目最後一項打錯,複製貼上的typo 感謝更正
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04/16 20:29, , 2F
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※ 編輯: FAlin (140.112.244.138), 04/16/2014 21:52:10
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04/16 22:15, , 6F
04/16 22:15, 6F
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中學
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