题目
题型:不详难度:来源:
Sn |
an-3 |
答案
∴6a2=4a1+2a3 ,即6q=4+2q2,解得 q=2.
∴an=2n-1,Sn=
1×(1-2n) |
1-2 |
Sn |
2n-1-3 |
2n-1 |
2n-1-3 |
5 |
2n-1-3 |
故答案为 7.
核心考点
举一反三
(1)求证:数列{an-2n}为等差数列;
(2)设数列{bn}满足bn=log2(an+1-n),若(1+
1 |
b2 |
1 |
b3 |
1 |
b4 |
1 |
bn |
n+1 |
a | 24 |
Sn |
an-3 |
1×(1-2n) |
1-2 |
Sn |
2n-1-3 |
2n-1 |
2n-1-3 |
5 |
2n-1-3 |
1 |
b2 |
1 |
b3 |
1 |
b4 |
1 |
bn |
n+1 |
a | 24 |