Question

已知f(n)=sinnπ/4,n∈Z,求f(1)+f(2)+…+f(2005)

Answer 1

这是有周期性的，把X＝1代进去，得到一个值，再继续代2，3，4… …直到你发现是几个一循环的就可以了，这一个循环里的各个值加起来，再用简单的周期性质就可以了

Answer 2

k∈Z
sin((8k+1)π/4)=sqrt(2)/2
sin((8k+2)π/4)=1
sin((8k+3)π/4)=sqrt(2)/2
sin((8k+4)π/4)=0
sin((8k+5)π/4)=-sqrt(2)/2
sin((8k+6)π/4)=-1
sin((8k+7)π/4)=-sqrt(2)/2
sin((8k+8)π/4)=0
sin((8k+1)π/4)+sin((8k+2)π/4)+...+sin((8k+8)π/4)=0;
2005=8*250+5;
f(1)+f(2)+…+f(2005)=f(1)+f(2)+f(3)+f(4)+f(5)=1+sqrt(2)/2=1.7071