已知等差数列{an}的公差d>0,其前n项和为Sn,若S3=12,且2a1,a2,1+a3成等比数列.
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已知等差数列{an}的公差d>0,其前n项和为Sn,若S3=12,且2a1,a2,1+a3成等比数列.
(Ⅰ)求{an}的通项公式;
(Ⅱ)记b
(Ⅰ)求{an}的通项公式;
(Ⅱ)记b
![已知等差数列{an}的公差d>0,其前n项和为Sn,若S3=12,且2a1,a2,1+a3成等比数列.](/uploads/image/z/15928348-4-8.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%85%AC%E5%B7%AEd%EF%BC%9E0%EF%BC%8C%E5%85%B6%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%EF%BC%8C%E8%8B%A5S3%3D12%EF%BC%8C%E4%B8%942a1%EF%BC%8Ca2%EF%BC%8C1%2Ba3%E6%88%90%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%EF%BC%8E)
(I)由题得:
2a1(a3+1)=a22
a1+a2+a3=12
即
a1(a1+2d+1) =8
a1+d=4,得d2+d-12=0.
∵d>0,∴d=3,a1=1.
∴{an}的通项公式an=1+3(n-1)=3n-2.
(II)∵bn=
1
an•an+1=
1
(3n-2)(3n+1)=
1
3(
1
3n-2-
1
3n+1).
∴Tn=b1+b2+b3+…+bn
=
1
3[(1-
1
4)+(
1
4-
1
7)+…+(
1
3n-2-
1
3n+1)]
=
1
3(1-
1
3n+1)
=
n
3n+1.
再问: 有详细解题过程吗
2a1(a3+1)=a22
a1+a2+a3=12
即
a1(a1+2d+1) =8
a1+d=4,得d2+d-12=0.
∵d>0,∴d=3,a1=1.
∴{an}的通项公式an=1+3(n-1)=3n-2.
(II)∵bn=
1
an•an+1=
1
(3n-2)(3n+1)=
1
3(
1
3n-2-
1
3n+1).
∴Tn=b1+b2+b3+…+bn
=
1
3[(1-
1
4)+(
1
4-
1
7)+…+(
1
3n-2-
1
3n+1)]
=
1
3(1-
1
3n+1)
=
n
3n+1.
再问: 有详细解题过程吗
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