[典例引領(lǐng)]
若各項(xiàng)均為正數(shù)的數(shù)列{an}的前n項(xiàng)和為Sn,且2=an+1 (n∈N*).
(1)求數(shù)列{an}的通項(xiàng)公式;
(2)若正項(xiàng)等比數(shù)列{bn},滿足b2=2,2b7+b8=b9,求Tn=a1b1+a2b2+…+anbn;
(3)對(duì)于(2)中的Tn,若對(duì)任意的n∈N*,不等式λ(-1)n<(Tn+21)恒成立,求實(shí)數(shù)λ的取值范圍.
解:(1)因?yàn)?/span>2=an+1,
所以4Sn=(an+1)2,且an>0,
則4a1=(a1+1)2,解得a1=1,
又4Sn+1=(an+1+1)2,
所以4an+1=4Sn+1-4Sn=(an+1+1)2-(an+1)2,
即(an+1-an-2)(an+1+an)=0,
因?yàn)?/span>an>0,所以an+1+an≠0,
所以an+1-an=2,
所以{an}是公差為2的等差數(shù)列,
又a1=1,
所以an=2n-1.
(2)設(shè)數(shù)列{bn}的公比為q,
因?yàn)?/span>2b7+b8=b9,所以2+q=q2,解得q=-1(舍去)或q=2,
由b2=2,得b1=1,故bn=2n-1.
因?yàn)?/span>Tn=a1b1+a2b2+…+anbn=1×1+3×2+5×22+…+(2n-1)×2