设数列{an}满足:a1=2,a2=5/3,an+2=5/3an+1+1/3an(n=1,2,3,...)(1)令bn=an+1-an(n=1,2,3,...),求数列{bn}的通项公式;(2)求数列{n×an}的前n项和Sn题意不清我修改一下题干不好意思:a1=2,a2=5/3,a(n+2)=5/3a(n+1)+1/3an
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/29 22:05:45
![设数列{an}满足:a1=2,a2=5/3,an+2=5/3an+1+1/3an(n=1,2,3,...)(1)令bn=an+1-an(n=1,2,3,...),求数列{bn}的通项公式;(2)求数列{n×an}的前n项和Sn题意不清我修改一下题干不好意思:a1=2,a2=5/3,a(n+2)=5/3a(n+1)+1/3an](/uploads/image/z/1765605-21-5.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97%EF%BD%9Ban%7D%E6%BB%A1%E8%B6%B3%EF%BC%9Aa1%3D2%2Ca2%3D5%2F3%2Can%2B2%3D5%2F3an%2B1%2B1%2F3an%28n%3D1%2C2%2C3%2C...%29%281%29%E4%BB%A4bn%3Dan%2B1-an%28n%3D1%2C2%2C3%2C...%29%2C%E6%B1%82%E6%95%B0%E5%88%97%EF%BD%9Bbn%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%9B%EF%BC%882%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97%7Bn%C3%97an%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%E9%A2%98%E6%84%8F%E4%B8%8D%E6%B8%85%E6%88%91%E4%BF%AE%E6%94%B9%E4%B8%80%E4%B8%8B%E9%A2%98%E5%B9%B2%E4%B8%8D%E5%A5%BD%E6%84%8F%E6%80%9D%EF%BC%9Aa1%3D2%2Ca2%3D5%2F3%2Ca%28n%2B2%29%3D5%2F3a%28n%2B1%29%2B1%2F3an)
设数列{an}满足:a1=2,a2=5/3,an+2=5/3an+1+1/3an(n=1,2,3,...)(1)令bn=an+1-an(n=1,2,3,...),求数列{bn}的通项公式;(2)求数列{n×an}的前n项和Sn题意不清我修改一下题干不好意思:a1=2,a2=5/3,a(n+2)=5/3a(n+1)+1/3an
设数列{an}满足:a1=2,a2=5/3,an+2=5/3an+1+1/3an(n=1,2,3,...)
(1)令bn=an+1-an(n=1,2,3,...),求数列{bn}的通项公式;
(2)求数列{n×an}的前n项和Sn
题意不清我修改一下题干不好意思:a1=2,a2=5/3,a(n+2)=5/3a(n+1)+1/3an(n=1,2,3,...)
设数列{an}满足:a1=2,a2=5/3,an+2=5/3an+1+1/3an(n=1,2,3,...)(1)令bn=an+1-an(n=1,2,3,...),求数列{bn}的通项公式;(2)求数列{n×an}的前n项和Sn题意不清我修改一下题干不好意思:a1=2,a2=5/3,a(n+2)=5/3a(n+1)+1/3an
给一个例题你看,思路一样的!
设a1=2,a2=5/3,an -2=5/3(an -1)-2/3(an)
求{nan}的前n项和Sn
递推公式是a(n-2)=5/3[a(n-1)]-2/3(an)
a(n-2)=5/3[a(n-1)-2/3(an)
3a(n-2)=5[a(n-1]-2(an)
2[an-a(n-1)]=3[a(n-1)-a(n-2)]
[an-a(n-1)]/[a(n-1)-a(n-2)]=3/2
所以{an-a(n-1)}是一个公比为3/2的等比数列,
an-a(n-1)=[a2-a1]*(3/2)^(n-2)=-1/3*(3/2)^(n-2)
an=8/3-(3/2)^(n-2)
所以nan=8n/3-n*(3/2)^(n-2).
Sn=4n(n+1)/3-8/3-(2n-4)*(3/2)^(n-1
请用括号表明你的各个量,比方说an+1 还是 a(n+1)?
5/3an是(5/3)an还是5/(3an)?