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※ 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 # -- 'You've sort of made up for it tonight,' said Harry. 'Getting the sword. Finishing the Horcrux. Saving my life.' 'That makes me sound a lot cooler then I was,' Ron mumbled. 'Stuff like that always sounds cooler then it really was,' said Harry. 'I've been trying to tell you that for years.' -- Harry Potter and the Deathly Hollows, P.308 -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 123.195.193.206 (臺灣) ※ 文章網址: https://www.ptt.cc/bbs/Math/M.1583603780.A.70A.html
chemmachine : 推 03/08 02:21
JI1 : michio kaku, liu zhiheng, timothy rainone 03/08 05:49
liang6159 : 謝謝 03/08 08:28
algebraic : 有趣 03/08 13:50