1 |
2 |
1 |
2 |
当公比不为1时,Sn=
1−qn |
1−q |
1 |
2 |
1−qn |
1−q |
1 |
2 |
1 |
2 |
1−qn+1 |
1−q |
1 |
2 |
∴
Sn+1+
| ||
Sn+
|
2qn+1+q−3 |
2qn+q−3 |
q(2qn+q−3)−q2+4q−3 |
2qn+q−3 |
−q2+4q−3 |
2qn+q−3 |
∵数列{Sn+
1 |
2 |
∴-q2+4q-3=0
∵q≠1,∴q=3
∴Sn=
1−qn |
1−q |
1−3n |
1−3 |
1 |
2 |
故答案为:
1 |
2 |
1 |
2 |
1 |
2 |
1 |
2 |
1−qn |
1−q |
1 |
2 |
1−qn |
1−q |
1 |
2 |
1 |
2 |
1−qn+1 |
1−q |
1 |
2 |
Sn+1+
| ||
Sn+
|
2qn+1+q−3 |
2qn+q−3 |
q(2qn+q−3)−q2+4q−3 |
2qn+q−3 |
−q2+4q−3 |
2qn+q−3 |
1 |
2 |
1−qn |
1−q |
1−3n |
1−3 |
1 |
2 |
1 |
2 |