Question

数列{An}的前n项和为Sn,A1=1,An+1=2Sn+1,

1.求{An}通项公式
2.等差数列{Bn}的各项均为正，其前几项的和为Tn,且T3=15,又a1+b1,a2+b2,a3+b3成等比，求Tn

Answer 1

A(n+1)=2Sn + 1.............1
An=2S(n-1) + 1..........2(n>=2)
1式减2 式得 A(n+1)-An=2An A(n+1)=3An A1=1
An=3^(n-1)

T3=b1+b2+b3=3b2=15 b2=5 b1=10-b3
a1+b1=11-b3 a2+b2=8 a3+b3=9+b3
8*8=(11-b3)(9+b3)
解得b3=7,-5(舍去）b1=3 bn=2n+1
Tn=(2+1)+(4+1)+(6+1)+...(2n+1)
=2(1+2+3+4+....+n)+n
=(1+n)n+n
=(n+2)n