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
(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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