Question

已知函数f(x)=cos派/2x,则f(1)+f(2)+...+f(2009)=?

Answer 1

当x=4kπ+1时，π/2x=2kπ+π/2 cosπ/2x=cos(2kπ+π/2)=0
当x=4kπ+2时，π/2x=2kπ+π cosπ/2x=cos(2kπ+π)=-1
当x=4kπ+3时，π/2x=2kπ+3π/2 cosπ/2x=cos(2kπ+3π/2)=0
当x=4kπ+4时，π/2x=2kπ+2π cosπ/2x=cos(2kπ+2π)=1
∴f(4k+1)+f(4k+2)+f(4k+3)+f(4k+4)=0
∴f(1)+f(2)+...+f(2009)=[f(1)+f(2)+f(3)+f(4)]+……+[f(2005)+f(2006)+f(2007)+f(2008)]+f(2009)
=f(2009)=cos(2009π/2) =cos(1004π+π/2)=cos(π/2)=0

Answer 2

f(x)=cos(πx/2)
T=4
f(1)=0,f(2)=-1,f(3)=0,f(4)=1,其和为0
f(1)+f(2)+...+f(2009)=[f(1)+f(2)+f(3)+f(4)]+[f(5)+f(6)+f(7)+f(8)]+...+[f(2005)+f(2006)+f(2007)+f(2008)]+f(2009)(共502个方括号）
=0+0+0+…+0+f(502×4+1)(共502个0）
=f(1)
=0