作者LPH66 (信じる力 奇跡起こすこと)
看板Math
標題Re: [中學]
時間Sun Mar 8 01:56:18 2020
※ 引述《liang6159 (liang6159)》之銘言：
: 若a^101=1,a不等於1,
: 求：（a^2/a-1)+(a^4/a^2-1)+(a^6/a^3-1)+....+(a^200/a^100-1)=?
: 這是JHMC題目，手機排版稍亂請見諒！
: → algebraic : 請問減一是在分母嗎 03/08 01:37
假設題目是這樣
首尾配對, 對首末兩項有
a^2/(a-1) + a^200/(a^100-1)
= (a + 1 + 1/(a-1)) + (a^100 + 1 + 1/(a^100-1))
= a + a^100 + 2 + 1/(a-1) + 1/(a^100-1) (x^2/(x-1) = x + 1 + 1/(x-1) 長除法)
= a + a^100 + 2 + (a-1+a^100-1)/[(a-1)(a^100-1)] (通分)
= a + a^100 + 2 + (a + a^100 - 2)/(a^101 - a - a^100 + 1)
= a + a^100 + 2 + (a + a^100 - 2)/(2 - a - a^100)
= a + a^100 + 2 - 1 = a + a^100 + 1
其他項類推, 由此原式 = a + a^2 + a^3 + ... + a^100 + 50
再由 a^101 = 1 得 a^101 - 1 = 0, 即 (a-1)(a^100+a^99+...+a+1) = 0
a 不為 1 故後一乘數為 0, 即 a^100+a^99+...+a+1 = 0, 於是原式 = 49 #
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1F：→ chemmachine : 推 03/08 02:21
2F：推 JI1 : michio kaku, liu zhiheng, timothy rainone 03/08 05:49
3F：推 liang6159 : 謝謝 03/08 08:28
4F：推 algebraic : 有趣 03/08 13:50