Question

已知f(n)=cosnπ/4,求f(1)+f(2)+...+f(100)的值？

Answer 1

函如启数f(n)=cosnπ/4的周期：
经观察,知f(1)=f(9)=f(17)=.=f(89)=cos(π/4)=√2/2
f(2)=f(10)=f(18)=.=f(90)=cos(2π/渣漏如4)=0
f(3)=f(11)=f(19)=.=f(91)=cos(3π/4)=-√2/2
f(4)=f(12)=f(20)=.=f(92)=cos(4π/4)=-1
f(5)=-f(1)
f(6)=-f(2)
f(7)=-f(3)
f(8)=-f(4)
故：f(1)+f(2)+...+f(100)=f(97)+f(98)+f(99)+f(100)=f(1)+f(2)+f(3)+f(4)=-1,10,已知f(n)=cosnπ/4,求f(1)+f(2)+...+f(100)的搜迟值
通俗易懂一点,