已知:f(n)=sin(nπ/4),求:f(1)+f(2)+…+f(100).

Question

已知:f(n)=sin(nπ/4),求:f(1)+f(2)+…+f(100).

Answer 1

f(1)=sin(π/4)=1/根号2
f(2)=sin(π/2)=1
f(3)=sin(3π/4)=1/根号2
f(4)=sin(π)=0
当n大等于5时,显然f(n)=-f(n-4)
所以
`f(1)+f(2)+…+f(100)
=(f(1)+...+f(8))+(f(9)+...+f(16))+...+f(97)+f(98)+f(99)+f(100)
=0+0+0...+f(1)+f(2)+f(3)+f(4)
=1+根号2