已知数列{an}满足:a1=1,a2=a(a>0).正项数列{bn}满足bn2=anan+1(n∈N*).若 {bn}是公比为
(1)求{an}的通项公式; (2)若a=
|
(1)
=bn+12 bn2
=an+1an+2 anan+1
=2,an+2 an
又∵a1=1,a2=a(a>0),
∴an=
.(
)n-1,n为正奇数2 a(
)n-2,n为正偶数2
(2)若a=
,则an=(2
)n-1(n∈N*),则{an}为等比数列,公比为2
,2
所以Sn=
=1×[1-(
)n]2 1- 2
.1-(
)n2 1- 2
Tn=
=17Sn-S2n an+1
[(1 1- 2
)n+2
-17]≤16 (
)n2
(8-17)=9(1 1- 2
+1).2
等号当且仅当(
)n=2
,即n=4时取到,16 (
)n2
n0=4.