Question

一道简单高一数列求和问题数列{AN}前N项和SN=3AN+1,则通项公式AN=?(N皆为角标)

Answer 1

Sn=3[Sn-S(n-1)]+1

2Sn=3S(n-1)-1

2Sn-2=3S(n-1)-3

所以(Sn-1)*[S(n-1)-1]=3/2

Sn-1是等比数列,q=3/2

S1=a1

S1=3S1+1

S1=-1/2

S1-1=-3/2

Sn=(-3/2)*(3/2)^(n-1)=-(3/2)^n

S(n-1)=-(3/2)^(n-1)

an=Sn-S(n-1)=(-3/2)*(3/2)^(n-1)-[-(3/2)^(n-1)]=-(1/2)*(3/2)^(n-1)