题目
题型:不详难度:来源:
A.130 | B.170 | C.210 | D.260 |
答案
所以由等差数列性质可得:sn,s2n-sn,s3n-s2n…为等差数列.
即30,100-30,S3n-100是等差数列,
∴2×70=30+S3n-100,解得S3n=210,
故选C.
核心考点
举一反三
A.若n≥2且an+1+an-1=2an,则{an}是等差数列 |
B.设数列{an}的前n项和为Sn,且2Sn=1+an,则数列{an}的通项an=(-1)n-1 |
C.若n≥2且an+1an-1=an2,则{an}是等比数列 |
D.若{an}是等比数列,且m,n,k∈N+,m+n=2k,则aman=ak2 |
Sn |
an-3 |
(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 |