Find the sum of the geometric series of 5+y+(y^2/3)+(y^3/9)+...
I know the a=5, and the r = y/3, but the sum isn t 5/(1-y/3). Please help
Geometric series help!?
2016-04-13 9:19 pm
回答 (1)
2016-04-13 9:26 pm
It's not a geometric series:
y/5 =/= (y^2/3)/y = y/3
you can split it up:
2 + (3 + y + y^2/3 + y^3/9 + ..)
and then the bit in brackets is geometric, the sum is then:
2 + 3/(1-(y/3))
= 2 + 9/(3-y)
= (15 - 2y)/(3 - y)
y/5 =/= (y^2/3)/y = y/3
you can split it up:
2 + (3 + y + y^2/3 + y^3/9 + ..)
and then the bit in brackets is geometric, the sum is then:
2 + 3/(1-(y/3))
= 2 + 9/(3-y)
= (15 - 2y)/(3 - y)
收錄日期: 2021-04-21 17:59:53
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160413131933AAp2UNP