问题
解答题
已知等差数列{an}的首项a1=1,公差d>0,且a2,a5,a14成等比数列. (1)求数列{an}的通项公式; (2)设bn=
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答案
(1)∵等差数列{an}的首项a1=1,公差d>0,且a2,a5,a14成等比数列,
∴(a1+d)(a1+13d)=(a1+4d)2
整理得:2a1d=d2,
∵a1=1,解得d=2(d=0舍去)
∴an=2n-1(n∈N*),
(2)bn=
=1 n(an+3)
=1 2n(n+1)
(1 2
-1 n
),1 n+1
∴Sn=b1+b2+…+bn
=
[(1-1 2
)+(1 2
-1 2
)+…+(1 3
-1 n
)]1 n+1
=
(1-1 2
),1 n+1
∴当n=1时,Sn取最小值S1=
(1-1 2
)=1 2
>1 4
.1 36
∴Sn>
.1 36