1) a(n+1) = 2Sn +1 [1]
a(n+2) = 2S(n+1) +1 [2]
[2]-[1]:a(n+2) - a(n+1) = 2*[S(n+1)-Sn]=2*a(n+1)
a(n+2)=3a(n+1)
所以 {an}是公比為3的等比數(shù)列,首項(xiàng)是1,an=a1*3^(n-1)=3^(n-1)
(2)
由上面的通項(xiàng)公式得到：a2=3,a3=9
因?yàn)閧bn}為等差數(shù)列,且T3＝15,所以b2=5,b1+b3=10
a1+b1=1+b1=1+10-b3=11-b3
a2+b2=3+5=8
a3+b3=9+b3
a1+b1,a2+b2,a3+b3成等比數(shù)列,所以 8*8=(11-b3)*(9+b3)
解此方程,得到：b3＝7,所以b1=10-7=3
bn=3+(n-1)*2= 2n+1
Tn= b1+b2+...+bn= 2*(1+2+...n)+n = n(n+1)+n=n(n+2)