题目

已知数列{an}中，an＞0，前n项的和为Sn，且满足Sn=(an+2)2.若数列{bn}满足bn=(t-1)(t＞1),Tn为数列{bn}的前n项的和，求Tn. 答案：解：∵Sn=(an+2)2,∴Sn-1=(an-1+2)2.∴Sn-Sn-1=(an+2)2-(an-1+2)2.∴Sn-Sn-1=(an+an-1+4)(an-an-1).∴8an=(an+an-1+4)(an-an-1).∴8an=an2-an-12+4an-4an-1.∴an2-an-12-4an-4an-1=0.∴(an-an-1-4)(an+an-1)=0.∵an＞0,∴an-an-1-4=0.∴an-an-1=4.∴数列{an}是公差为4的等差数列.∵a1=(a1+2)2,∴a1=2.∴an=a1+(n-1)×4=4n-2.∴bn==(t-1)n.∴Tn=(t-1)(t-1)(t≠2),Tn=n(t=2).∴Tn=(t-1).当1＜t＜2时,Tn=;当t≥2时,无极限.

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