∴Sn-1=n2-n,(n≥2),
∴Sn=(n+1)2-(n+1)=n2+n(n≥1)
∴当n=1时,a1=S1=2;
当n≥2时,an=Sn-Sn-1=n2+n-(n-1)2-(n-1)=2n,
所以数列{an}的通项公式为an=2n(n∈N*).
( II)当n≥2时,an=Sn-Sn-1
=kan-kan-1+2n-2,
∴(k-1)an=kan-1-2n+2,a1=S1=ka1,
若k=1,则an-2n-1=-1,
从而{an-2n-1}为公比为1的等比数列,不合题意;
若k≠1,则a1=0,a2=
2 |
1-k |
4-6k |
(1-k)2 |
5k-3 |
1-k |
-7k2+8k-3 |
(k-1)2 |
由题意得,(a2-5)2=(a1-3)(a3-7)≠0,
∴k=0或k=
3 |
2 |
当k=0时,Sn=n2-n,an=2n-2,an-2n-1=-3,不合题意;
当k=
3 |
2 |
∵a1-2×1-1=-3≠0,an-2n-1≠0,{an-2n-1}为公比为3的等比数列,
∴an-2n-1=-3n,
∴an=2n-3n+1,
∴Sn=n2+2n-
3n+1 |
2 |
3 |
2 |