Question

等比数列

设数列{an}的前n项和为Sn,已知a1=1,Sn-1=4an+2,求an通项公式

Answer 1

A1+A2=S2=4A1+2
A2=3A1+2=5
S(n+1)=4An+2
Sn=4A(n-1)+2
A(n+1)=S(n+1)-Sn=4An-4A(n-1)
A(n+1)-2An=2An-4A(n-1)=2(An-2A(n-1))
A2-2A1=5-2=3
{A(n+1)-2An}是以3为首项，2为公比的等比数列
A(n+1)-2An=3×2^(n-1)
An-2A(n-1)=3×2^(n-2)
An/2^n-2A(n-1)/2^n=3×2^(n-2)/2^n
An/2^n-A(n-1)/2^(n-1)=3/4
A1/2^1=1/2
{An/2^n}是以1/2为首项，3/4为公差的等差数列
An/2^n=1/2+(n-1)3/4=(3n-1)/4
An=(3n-1)×2^(n-2)

Answer 2