If a(1)=4, a(2)=5 and a(n+2)=a(n)+a(n+1) for any positive integer n, then a(10)=?
Please show the steps.
更新1:
To Kitty Fung , Is this the only method???
更新2:
This method seems too simple
Arithmetic Sequences plz help
2013-04-14 3:07 am
Let a(n) be the n th term of a sequence.
If a(1)=4, a(2)=5 and a(n+2)=a(n)+a(n+1) for any positive integer n, then a(10)=?
Please show the steps.
If a(1)=4, a(2)=5 and a(n+2)=a(n)+a(n+1) for any positive integer n, then a(10)=?
Please show the steps.
回答 (2)
2013-04-14 3:37 am
✔ 最佳答案
a(1)=4a(2)=5
a(3)=a(1)+a(2)=4+5=9
a(4)=a(2)+a(3)=5+9=14
a(5)=a(3)+a(4)=9+14=23
a(6)=a(4)+a(5)=14+23=37
a(7)=a(5)+a(6)=23+37=60
a(8)=a(6)+a(7)=37+60=97
a(9)=a(7)+a(8)=60+97=157
a(10)=a(8)+a(9)=97+157=254
Therefore, a(10)=254
2013-04-14 13:20:19 補充：
This should be the only method.
2013-04-15 4:48 am
By deduction, we have a(10) = a(9) + a(8) = [ a(8) + a(7) ] + a(8) .... = 2a(8) + a(7) ... = 3a(7) + 2a(6) ... = 5a(6) + 3a(5) = 8a(5) + 5a(4) = 13a(4) + 8a(3) = 21a(3) + 13a(2) = 34a(2) + 21a(1) = 34(5) + 21(4) = 254
收錄日期: 2021-04-23 19:54:51
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130413000051KK00224