數學一問(展開 項數 次方 )求解答 20點

(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 !

x^100-1=0
實數real number 的話,答案是2個
虛數complex number 的話,是100 個

知道啦
我以後會緊樣表示
唔該曬︿ ︿
_

次方可以^ 表示
例如x的二次方為 x^2