设数列{an}满足:a1=2,a2=5/3,an+2=5/3an+1+1/3an(n=1,2,3,...)
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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,...)
(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,...)
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
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
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