已知数列｛an｝是各项均不为0的等差数列,公差为d,Sn为其前n项和,且满足an^2=S2n-1,n∈N*,数列｛bn｝满足bn=1/(an·an+1),Tn为数列｛bn｝的前n项和.(1)求a1、d、和Tn(2)若对任意的n属于N*,不等式λTn

已知数列｛an｝是各项均不为0的等差数列,公差为d,Sn为其前n项和,且满足an^2=S2n-1,n∈N*,数列｛bn｝满足bn=1/(an·an+1),Tn为数列｛bn｝的前n项和.(1)求a1、d、和Tn(2)若对任意的n属于N*,不等式λTn
已知数列｛an｝是各项均不为0的等差数列,公差为d,Sn为其前n项和,且满足an^2=S2n-1,n∈N*,数列｛bn｝满
足bn=1/(an·an+1),Tn为数列｛bn｝的前n项和.
(1)求a1、d、和Tn
(2)若对任意的n属于N*,不等式λTn(3)是否存在正整数m,n(1

已知数列｛an｝是各项均不为0的等差数列,公差为d,Sn为其前n项和,且满足an^2=S2n-1,n∈N*,数列｛bn｝满足bn=1/(an·an+1),Tn为数列｛bn｝的前n项和.(1)求a1、d、和Tn(2)若对任意的n属于N*,不等式λTn
an²=S2n-1
a1²=S1
a1=1、0
数列｛an｝是各项均不为0的=>
a1=1
an²=S2n-1
a2²=S3=a3+a2+a1
(a1+d)²=a1+2d+a1+d+a1
d²+2d+1=3d+3
d²-d-2=0
d=2、-1
数列｛an｝是各项均不为0的=>a2!=1-1=>
d=2
bn=1/(an·an+1)(n+1是角标吧?不是就别往下看了-.-b)
=1/[(2n-1)·(2n+1)]
=[1/(2n-1) - 1/(2n+1)]/2
Tn=1/2 · [1/1-1/3 + 1/3-1/5 +...+1/(2n-1) - 1/(2n+1)]
=1/2 · [1 - 1/(2n+1)]
=n/(2n+1)
(2):
λTn

Sn=na1+[n(n-1)d]/2,所以an²=S(2n-1)=(2n-1)a1+(n-1)(2n-1)d,化简得到an=0,或者an=2n-1,如题an均不为0，所以an=1,且d=2.