已知数列{an}中,a2=2,前n项和为Sn,且Sn=n(an+1)/2证明数列{an+1-an}是等差数列

来源：学生作业帮编辑：拍题作业网作业帮分类：综合作业时间：2024/04/25 14:43:44

Sn=n(an+1)/2
S(n+1)=(n+1)[a(n+1)+1]/2
用下式减上式
a(n+1)=[(n+1)a(n+1)-nan+1]/2
即2a(n+1)=[(n+1)a(n+1)-nan+1]
即(n-1)a(n+1)-nan+1=0
即(n-1)a(n+1)-(n-1)-[nan-n]=0
即[a(n+1)-1](n-1)=(an-1)n
即[a(n+1)-1]/n=(an-1)/(n-1)
这样就得到了一个递推式
[a(n+1)-1]/n=(an-1)/(n-1)＝[a(n-1)-1]/(n-2)
=.
=(a2-1)/1=1
所以an-1=n-1
所以an=n

已知数列{an}中,a2=2,前n项和为Sn,且Sn=n(an+1)/2证明数列{an+1-an}是等差数列数列{an}中,a2=2,前n项和为sn,且sn=n(an+1)/2 证明{a-an}是等差数列数列{an}前n项和为Sn,且an+Sn=-2n-1 证明{an+2}是等比数列数列{an}前n项和Sn=npa[n](n是正整数),且a1不等于a2,(1)求p的值(2)证明{an}为等差数列已知数列{an}的前n项和为Sn,且满足an+2Sn*Sn-1=0,a1=1/2.求证：{1/Sn}是等差数列已知正项数列an的前n项和为Sn,a1=1,（an-2）²=8Sn-1.证明an是等差数列. 数列An的前n项和为Sn,已知A1=1,An+1=Sn*(n+2)/n,证明数列Sn/n是等比数列已知数列an的前n项和为sn,且sn+an=n^2+3n+5/2,证明数列{an-n}是等比数列已知数列{An}的前n项和Sn=3n²-2n,证明数列{An}为等差数列数列{an}前n项和为Sn,且Sn=n-5an-85,证明{an-1}是等比数列已知数列{An},Sn是其前n项和,且满足3An=2Sn+n,n为正整数,求证数列{An+1/2}为等比数列已知数列{an}的前n项和为Sn,且满足Sn=Sn-1/2Sn-1 +1,a1=2,求证{1/Sn}是等差数列