1.求等比數列的無限項之和
-1/2，1/6，-1/18，1/54，....
2.已知等比數列的首項和公比分別是k和k/5，且無限項之和是20。求k
的值。
3.若等比數列40，40R，40R2，...的無限項之和是60，求公比。
關於等比數列的一個問題
2011-09-17 10:34 pm
回答 (2)
2011-09-18 9:53 pm
✔ 最佳答案
1.(-1/2)/(1-(-1/3))
=(-1/2)/(1+1/3)
=(-1/2)/(4/3)
=(-1/2) x (3/4)
=-3/8
2.
k/(1-(k/5)=20
k=20-(20k)/5
5k=100-20k
25k=100
k=4
3.
40/(1-R)=60
40=60-60R
R=1/3
2011-09-17 10:59 pm
1 等比數列的無限項之和
= a/(1 - r)
= (-1/2)/(1 + 1/3)
= -3/8
2 k/(1 - k/5) = 20
k = 20(1 - k/5)
k = 20 - 4k
k = 4
3 40/(1 - R) = 60
2/3 = 1 - R
R = 1/3
= a/(1 - r)
= (-1/2)/(1 + 1/3)
= -3/8
2 k/(1 - k/5) = 20
k = 20(1 - k/5)
k = 20 - 4k
k = 4
3 40/(1 - R) = 60
2/3 = 1 - R
R = 1/3
收錄日期: 2021-04-26 14:43:58
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110917000051KK00481