已知等差数列{an}前三项的和为-3,前三项的积为8. (1)若a2,a3,a1成等比数列,求数列{|an|}的前n项和. (2)若a2,a3,a1不成等比数列,求数列{
|
(1)设等差数列{an}的公差为d,
由题意得
解得3a1+3d=-3 a1(a1+d)(a1+2d)=8
或a1=2 d=-3
.a1=-4 d=3
∴an=2-3(n-1)=-3n+5或an=-4+3(n-1)=3n-7.
当an=3n-7时,a2,a3,a1分别为-1,2,-4,成等比数列,满足条件.
设数列{|an|}的前n项和为Sn.
∴当n=1,2时,|an|=7-3n,Sn=
=-n(4+7-3n) 2
n2+3 2
n;11 2
当n≥3时,|an|=3n-7,
Sn=-a1-a2+a3+a4+…+an
=5+(n-2)(2+3n-7) 2
=
n2-3 2
n+10,11 2
综上可得:|an|=|7-3n|=-3n+7,n=1,2 3n-7,n≥3
Sn=-
n2+3 2
n,n=1,211 2
n2-3 2
n+10,n≥311 2
(2)当an=-3n+5时,a2,a3,a1分别为-1,-4,2,不成等比数列.
=1 anan+1
=1 (3n-5)(3n-2)
(1 3
-1 3n-5
),1 3n-2
∴Tn=
[(-1 3
-1)+(1-1 2
)+…+(1 4
-1 3n-5
)]1 3n-2
=
[-1 3
-1 2
]1 3n-2
=
.n -6n+4