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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.

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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
S
(Ⅰ)设公差为d,由条件得

5a1+
5×4
2d=30
(a1+2d)2=a1(a1+8d),得a1=d=2.
∴an=2n
Sn=2n+
n(n-1)×2
2=n2+n;
(Ⅱ)∵
1
Sn+an+2=
1
n2+n+2n+2=
1
n2+3n+2
=
1
(n+1)(n+2)=
1
n+1-
1
n+2.

1
S1+a1+2+
1
S2+a2+2+…+
1
Sn+an+2
=(
1
2-
1
3)+(
1
3-
1
4)+…+(
1
n+1-
1
n+2)
=
1
2-
1
n+2>
12
25.

1
n+2<
1
2-
12
25=
1
50,
即:n+2>50,n>48.
∴n的最小值为48.