Question

已知数列{An}中，A1=1.前n项和为Sn且Sn+1=3/2Sn＋1.求数列{An}的通项公式

Answer 1

S(n+1)=3/2Sn+1
S(n+1)+2=3/2Sn+3
S(n+1)+2=3/2(Sn+2)
[S(n+1)+2]/[(Sn+2)]=3/2
所以Sn+2是以3/2为公比的等比数列
Sn+2=(S1+2)*q^(n-1)
Sn+2=(a1+2)*q^(n-1)
Sn+2=(1+2)*(3/2)^(n-1)
Sn=3*(3/2)^(n-1)-2

Sn=3*(3/2)^(n-1)-2
S(n-1)=3*(3/2)^(n-2)-2
an=Sn-S(n-1)
=3*(3/2)^(n-1)-2-3*(3/2)^(n-2)+2
=3*3/2*(3/2)^(n-2)-3*(3/2)^(n-2)
=(9/2-3)*(3/2)^(n-2)
=3/2*(3/2)^(n-2)
=(3/2)^(n-1)

Answer 2

Sn=3/2S(n-1)+1
S(n-1)=3/2S(n-2)+1
两式相减，得
An=3/2A(n-1)
所以数列为首项为1，公比为3/2的等比数列
An=(3/2)的n-1次方