数列{an}的前n项和为Sn满足:Sn=2n2-3n+1,则a4+a5+…+a10等于
[ ]
A.
171
B.
21
C.
10
D.
161
答案:D
解析:
解析:
|
因a4+a5+…+a10=S10-S3=161. |
数列{an}的前n项和为Sn满足:Sn=2n2-3n+1,则a4+a5+…+a10等于
171
21
10
161
|
因a4+a5+…+a10=S10-S3=161. |