1.(2015大连一模)等差数列{an}的前n项和为Sn,等比数列{bn}的公比为,满足S3=15,a1+2b1=3,a2+4b2=6.
(1)求数列{an},{bn}的通项公式an,bn;
(2)求数列{an·bn}的前n项和Tn.
解:(1)设{an}的公差为d,
所以
解得a1=2,d=3,b1=,
所以an=3n-1,bn=.
(2)由(1)知Tn=2×+5×+8×+…+(3n-4)·+(3n-1), ①
①×Tn=2×+5×+…+(3n-4)×+(3n-1), ②
①-②得
Tn=2×+3×-(3n-1)·=1+3×-(3n-1)·,
