在数列an中,a1=1,an+1=3an,等差数列bn各项均为正数,前n项和为Tn,T3=15,a1+b1,a2+b2,a3+b3成等比,求Tn
问题描述：
在数列an中,a1=1,an+1=3an,等差数列bn各项均为正数,前n项和为Tn,T3=15,a1+b1,a2+b2,a3+b3成等比,求Tn
答
a(n+1)=3a(n),
{a(n)}是首项为1,公比为3的等比数列.
b(n)=b+(n-1)d. T(n)=nb + n(n-1)d/2.
15=T(3)= 3b + 3d, 5=b+d.
[3+b+d]^2 = [a(2)+b(2)]^2 = [1+b][9+b+2d],
[3+5]^2 = [1+5-d][9+5+d],
64=(6-d)(14+d)=84-8d-d^2,
0=d^2+8d-20=(d+10)(d-2), d>0. d=2. b = b+d-d=5-d=3.
T(n)=nb+n(n-1)d/2 = n^2 +2n
答
a(n+1)=3a(n),{a(n)}是首项为1,公比为3的等比数列.a(n)=3^(n-1).b(n)=b+(n-1)d.T(n)=nb + n(n-1)d/2.15=T(3)= 3b + 3d,5=b+d.[3+b+d]^2 = [a(2)+b(2)]^2 = [a(1)+b(1)][a(3)+b(3)] = [1+b][9+b+2d],[3+5]^2 = [1+5-d...