求等比級數1-1/2+1/4-1/8+...............無限項之和
列式
求等比級數1-1/2+1/4-1/8+...............無限項之和
2007-01-02 2:49 am
回答 (3)
2007-01-02 2:53 am
✔ 最佳答案
1-1/2+1/4-1/8+...............首項a = 1
公比R = -1/2
用a/(1-R)
1-1/2+1/4-1/8+...............
= 1/[1-(-1/2)]
= 1/[1+1/2]
= 2/3.
2007-01-02 3:20 am
common rate=(-1/2)/1= -1/2
無限項之和=a/(1-r) (a=第一項,r=common rate)
無限項之和=1/(1+1/2)=1/(3/2)=2/3
so
無限項之和=2/3
無限項之和=a/(1-r) (a=第一項,r=common rate)
無限項之和=1/(1+1/2)=1/(3/2)=2/3
so
無限項之和=2/3
參考: 中五生的腦
2007-01-02 2:57 am
笫2頂=1/4-1/8=1/8
笫1頂=1-1/2=1/2 公比=1/8 / 1/2=1/4
s(無限)= 笫一頂/ 1-公比
= 1/2 / 1-1/4
= 1/2 / 3/4
=2/3
笫1頂=1-1/2=1/2 公比=1/8 / 1/2=1/4
s(無限)= 笫一頂/ 1-公比
= 1/2 / 1-1/4
= 1/2 / 3/4
=2/3
收錄日期: 2021-04-13 16:02:25
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070101000051KK04050