(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. 
(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. 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.