S(n+1)=4(An)+2
Sn=4A(n-1)+2
兩式相減
A(n+1)=S(n+1)-Sn=4An-4A(n-1)
A(n+1)-4An+4A(n-1)=0
A(n+1)-2An=2An-4A(n-1)=2(An-2A(n-1))
S2=4A1+2=4+2=6
A2=S2-A1=6-1=5
A2-2A1=5-2=3
{A(n+1)-2An},即｛bn}是以3為首項(xiàng),2為公比的等比數(shù)列
A(n+1)-2An=3×2^(n-1)
兩邊同除2^(n+1)
A(n+1)/2^(n+1)-2An/2^(n+1)=3×2^(n-1)/2^(n+1)
A(n+1)/2^(n+1)-An/2^n=3/4
依此類推
An/2^n-A(n-1)/2^(n-1)=3/4
A(n-1)/2^(n-1)-A(n-2)/2^(n-2)=3/4
……
A2/2-A1/1=3/4
上式相加,相同項(xiàng)消去
An/2^n-A1/2^1=3(n-1)/4
An/2^n=3(n-1)/4+1/2=(3n-1)/4
An=(3n-1)×2^(n-2)