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已知数列{an}是公差不为零的等差数列,其前n项和为Sn,且S5=30,又a1,a3,a9成等比数列.
(Ⅰ)求Sn
(Ⅱ)若对任意n>t,n∈N,都有
1
S1+a1+2
+
1
S2+a2+2
+…+
1
Sn+an+2
>
12
25
,求t的最小值.
人气:173 ℃ 时间:2019-08-19 03:28:26
解答
(Ⅰ)设公差为d,由条件得5a1+5×42d=30(a1+2d)2=a1(a1+8d),得a1=d=2.∴an=2n,Sn=2n+n(n-1)×22=n2+n;(Ⅱ)∵1Sn+an+2=1n2+n+2n+2=1n2+3n+2=1(n+1)(n+2)=1n+1-1n+2.∴1S1+a1+2+1S2+a2+2+…+1Sn+an+2=(12-13)+...
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