等差数列｛an｝满足:a2=5,a4+a10=30,｛an｝的前n项和为sn.(1)求an及Sn(2)数列｛bn｝满足bn(an^2---1)=8(n属于N+)数列bn前n项和为Tn,求证,Tn＜2

Question

等差数列｛an｝满足:a2=5 ,a4+a10=30,｛an｝的前n项和为sn.
(1)求an及Sn
(2)数列｛bn｝满足bn(an^2---1)=8(n属于N+)数列bn前n项和为Tn,求证,Tn＜2

Answer 1

(1) a2=a1+d=5,a4+a10=(a1+3d)+(a1+9d)=2a1+12d=30
∴a1=3,d=2,∴an=a1+(n-1)d=3+2(n-1)=2n+1
Sn=n(a1+an)/2=n(3+2n+1)/2=n^2+2n
∴an=2n+1,Sn=n^2+2n (n∈N+)
(2) bn=8/[(an)^2-1]=8/[(2n+1)^2-1]=8/[2n(2n+2)]=2/[n(n+1)]=2[1/n-1/(n+1)]
∴Tn=2[1-1/2+1/2-1/3+1/3-1/4+……+1/n-1/(n+1)]
=2[1-1/(n+1)]