题目
题型:不详难度:来源:
1 |
a |
1 |
a2 |
1 |
an-1 |
答案
1 |
a |
1 |
a2 |
1 |
an-1 |
将其每一项拆开再重新组合得Sn=(1+
1 |
a |
1 |
a2 |
1 |
an-1 |
当a=1时,Sn=n+
(3n-1)n |
2 |
(3n+1)n |
2 |
当a≠1时,Sn=
1-
| ||
1-
|
(3n-1)n |
2 |
a-a1-n |
a-1 |
(3n-1)n |
2 |
核心考点
举一反三
(1)求数列{an}的通项公式an及前n项和Sn;
(2)若数列{bn}满足bn+1-bn=an(n∈N*),且b1=3,求数列{
1 |
bn-n |
(1)求数列{an}的通项an;
(2)若数列{bn}是等差数列,且b1=a1,bm=am,判断数列{an}前m项的和Sm与数列{bn-
1 |
2 |