數學!!急數列

👨🏻‍🏫 學術台數學

2012-10-20 5:52 am

a1=1
a2=-1
a3=1
a4=-1......
an=?

a1=1 over 2‧5
a2=1 over 5‧8
a3=1 over 8‧11
a4=1 over 11‧14......
an=?

a1=-1
a2=3
a3=-5
a4=7
an=?

回答 (3)

2012-10-20 6:47 am

✔ 最佳答案

an= (-1)^(n+1)

an= 1/((3n-1) x (3n+2))

an= (-1)^n x (2n-1)

參考: me

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2012-10-20 6:23 am

Do you want a_n expressed in n or a_(n-1) ?

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2012-10-20 5:53 am

1.1
other i don't know

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收錄日期: 2021-04-13 19:03:36
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20121019000051KK00555

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