Question

已知数列{an}的前n项和为sn,且2sn=3an-2n,（n属于N*）,求证：数列{1+an}是...？

Answer 1

2sn=3an-2n2s(n-1)=3a(n-1)-2(n-1)2(sn-s(n-1))=3an-2n-3a(n-1)+2n-42an=3an-3a(n-1)-43a(n-1)+3=an+1(an+1)/(a(n-1)+1)=32s1=3a1-2a1=2a1+1=3所以数列{1+an}为以3为首项,3为等比的等比数列.an+1=3*3(n-1)=3^nan=3^n...,6,2S(n)=3a(n)-2n,
2a(n+1)=2S(n+1)-2S(n)=3a(n+1)-2(n+1)-[3a(n)-2n]=3a(n+1)-3a(n)-2,
0=a(n+1)-3a(n)-2=[a(n+1)+1] - 3[a(n)+1],
[a(n+1)+1]=3[a(n)+1],
{a(n)+1}是首项为a(1)+1,公比为3的等比数列.
2a1=2S1=3a1-2
a1=2
an+1=(a1+1)*q^(n-1)=3*3^(n-1)=3^n
an=3^n-1,2,已知数列{an}的前n项和为sn,且2sn=3an-2n,（n属于N*）,求证：数列{1+an}是...
已知数列{an}的前n项和为sn,且2sn=3an-2n,（n属于N*）,求证：数列{1+an}是等比数列,并求数列{an}的通向公式an