问题
解答题
已知等差数列{an}满足:a2+a4=14,a6=13,{an}的前n项和为Sn. (Ⅰ)求an及Sn; (Ⅱ)令bn=
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答案
(I)设首项为a1,公差为d,则
∵a2+a4=14,a6=13,∴2a1+4d=14 a1+5d=13
∴a1=3,d=2
∴an=3+2(n-1)=2n+1,Sn=3n+
×2=n2+2n;n(n-1) 2
(Ⅱ)证明:bn=
=1 an2-1
(1 4
-1 n
)1 n+1
∴Tn=
(1-1 4
+1 2
-1 2
+…+1 3
-1 n
)=1 n+1
(1-1 4
)<1 n+1 1 4
∵Tn单调递增,∴Tn≥T1=1 8
∴
≤Tn<1 8
.1 4