例1 已知{an}為等差數(shù)列,前n項(xiàng)和為Sn(n∈N*),{bn}是首項(xiàng)為2的等比數(shù)列,且公比大于0,b2+b3=12,b3=a4-2a1,S11=11b4.
(1)求{an}和{bn}的通項(xiàng)公式;
(2)求數(shù)列{a2nb2n-1}的前n項(xiàng)和(n∈N*).
解 (1)設(shè)等差數(shù)列{an}的公差為d,等比數(shù)列{bn}的公比為q.
由已知b2+b3=12,得b1(q+q2)=12,
而b1=2,所以q2+q-6=0.
又因?yàn)?i>q>0,解得q=2,所以bn=2n.
由b3=a4-2a1,可得3d-a1=8①,由S11=11b4,可得a1+5d=16②,聯(lián)立①②,
解得a1=1,d=3,由此可得an=3n-2.
所以數(shù)列{an}的通項(xiàng)公式為an=3n-2,數(shù)列{bn}的通項(xiàng)公式為bn=2n.
(2)設(shè)數(shù)列{a2nb2n-1}的前n項(xiàng)和為Tn,由a2n=6n-2,b2n-1=2×4n-1,有a2nb2n-1=(3n-1)×4n,故
Tn=2×4+5×42+8×43+…+(3n-1)×4n,
4Tn=2×42+5×43+8×44+…+(3n-4)×4n+(3n-1)×4n+1,