题目

f(0)=cos(npi/2 + pi/4),求f(1)+f(2)+.+f(2002)=?

提问时间：2021-02-25

答案

f(n)=cos(nπ/2 + π/4)
= cos[π(2n+1)/4]
当n=1时,π(2n+1)/4=3π/4
当n=2时,π(2n+1)/4=5π/4
当n=3时,π(2n+1)/4=7π/4
当n=4时,π(2n+1)/4=9π/4
当n=5时,π(2n+1)/4=11π/4
当n=6时,π(2n+1)/4=13π/4
∵cos[3π/4]+cos[7π/4]=0
cos[5π/4]+cos[9π/4]=0
cos[11π/4]+cos[15π/4]=0
.
∴它的周期是T=4
2002/4的商是500余数是2
∴f(1)+f(2)+.+f(2002)
=cos(3π/4)+cos(5π/4)+cos(7π/4)+...cos(2002π/4)
=(0)的200次方+cos[3π/4]+cos[7π/4]=0