已知{an}是等差数列,其前n项和为Sn,{bn}是等比数列,且a1=b1=2,a4+b4=27,S4-b4=10. (1)求数列{an}与{bn}的通项公式; (2)记Tn=a1b1+a2b2+…+anbn,n∈N*,证明:Tn-8=an

问题描述:

已知{an}是等差数列,其前n项和为Sn,{bn}是等比数列,且a1=b1=2,a4+b4=27,S4-b4=10.
(1)求数列{an}与{bn}的通项公式;
(2)记Tn=a1b1+a2b2+…+anbn,n∈N*,证明:Tn-8=an-1bn+1(n∈N*,n≥2).

(1)设等差数列的公差为d,等比数列的公比为q,
由a1=b1=2,得a4=2+3d,b4=2q3,s4=8+6d,
由a4+b4=27,S4-b4=10,得方程组

2+3d+2q3=27
8+6d−2q3=10

解得
d=3
q=2

所以:an=3n-1,bn=2n
(2)证明:由第一问得:Tn=2×2+5×22+8×23+…+(3n-1)×2n;   ①;
2Tn=2×22+5×23+…+(3n-4)×2n+(3n-1)×2n+1,②.
由①-②得,-Tn=2×2+3×22+3×23+…+3×2n-(3n-1)×2n+1
=
6×(1−2n)
1−2
-(3n-1)×2n+1-2
=-(3n-4)×2n+1-8.
即Tn-8=(3n-4)×2n+1
而当n≥2时,an-1bn+1=(3n-4)×2n+1
∴Tn-8=an-1bn+1(n∈N*,n≥2).