Question

数列{a_n}的通项公式a_n=ncos

nπ

2

，其前n项和为S_n，则S₂₀₁₃=

1006

．

Answer 1

分析：算出a_4n+1+a_4n+2+a_4n+3+a_4n+4=2，于是S2013=

2012

4

×2+a2013即可得出．

解答：解：∵a4n+1=(4n+1)cos

(4n+1)π

2

=(4n+1)cos

π

2

=0，a4n+2=(4n+2)cos

(4n+2)π

2

=-（4n+2），
a4n+3=(4n+3)cos

(4n+3)π

2

=0，a4n+4=(4n+4)cos

(4n+4)π

2

=4n+4．
∴a_4n+1+a_4n+2+a_4n+3+a_4n+4=2，
于是S2013=

2012

4

×2+a2013=1006．
故答案为1006．

点评：正确找出其周期性是解题的关键．