Question

已知f(n)=sin nπ/4,n∈Z,则f(1)+f(2)+…+f(100)=?

需要详细步骤，谢谢！

Answer 1

f(n)=sin(nπ/4),n∈Z
最小正周期是T=2π/(π/4)=8

而一个周期内的和，根据对称性正负抵消为0

所以f(1)+f(2)+…+f(100)=f(1)+f(2)+f(3)+f(4)=sin(π/4)+sin(π/2)+sin(3π/4)+sin(π)=√2+1

Answer 2

sin(x+π)=-sinx
f(1)+…+f(100)=f(1)+f(2)+f(3)+f(4)=1+√2