數學一問(展開項數次方 )求解答 20點

2010-07-22 11:44 pm

若展開數式
99 98 97 96 2
(x-1)(x + x+x + x...............................+x+x+1)
共有多少項?簡述答案。

由於次方5g點打
99 啊 98啊個D係
第二個( )入面個X ge

更新1:

要簡述答案點解緊樣答

回答 (4)

用戶2053

2010-07-22 11:52 pm

✔ 最佳答案

(x - 1) (x^99 + x^98 + x^97 + x^96..............+ x^2 + x + 1)= x(x^99 + x^98 + x^97 + x^96..............+ x^2 + x + 1) - (x^99 + x^98 + x^97 + x^96..............+ x^2 + x + 1)= x^100 + x^99 + x^98 + x^97 + ...... + x^3 + x^2 + x) - (x^99 + x^98 + x^97 + x^96..............+ x^2 + x + 1) ............(x^99 + x^98 + x^97 + ...+ x^2 + x) 對消後得 := x^100 - 12項

2010-07-22 19:57:50 補充：
簡述答案 :

由等比級數求和公式 :

1 + x + x^2 + .... + x^98 + x^99

= 1(1 - x^100)/(1 - x) , (首項 = 1 , 公比 = x)

故(x - 1) (x^99 + x^98 + x^97 + x^96..............+ x^2 + x + 1)

= (x - 1) 1(1 - x^100)/(1 - x)

= x^100 - 1

Hope I can help you !

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