已知数列{an}中,a1=2,a2=4,a(n+2)=a(n+1)-an,则a10等于?
人气:476 ℃ 时间:2020-04-16 05:38:32
解答
由a(n+2)=a(n+1)-an,和a(n+3)=a(n+2)-a(n+1)
得an+a(n+3)=0,
所以an+a(n+9)=0,即有a1+a10=0,
所以a10=-2
推荐
- 已知数列{an}中,a1=2,an+1=an+2n(n∈N+)则a10等于
- 数列:已知A1=2,A2=1,且An分之1减A(n-1)分之1等于A(n+1)分之1减An,求A10
- 已知数列{an}中,a1=1,a2=2+3,a3=4+5+6,a4=7+8+9+10,…,则a10=( ) A.610 B.510 C.505 D.750
- 在数列an中,若a1=2,na(n+1)=(n+1)an+2(n属于N*),则a10等于
- 数列{an}满足a1=1,当n大于等于2时,an=an-1+n,则a10=多少?
- 英语高手请进!~ 帮我翻一个词语
- you should fight against u___________ treatment.
- We ____________(have)math on Monday.
猜你喜欢