热心网友
f=cos1+ 2cos2+ …… +ncosncos1cosk=1/2[cos(k+1)+cos(k-1)]==fcos1=cos1cos1+ 2cos1cos2+ 3cos1cos3+…… ++(n-2)cos1cos(n-2)++(n-1)cos1cos(n-1)+ncos1cosn==1/2[cos(1+1)+cos(1-1)]+2/2[cos(2+1)+cos(2-1)]+3/2[cos(3+1)+cos(3-1)]+…… +(n-2)/2[cos(n-1)+cos(n-3)]+(n-1)/2[cosn+(n-2)]+n/2[cos(n+1)+cos(n-1)]==1/2+cos1+2cos2+…… +ncosn-(n+1)/2[cosn]+n/2[cos(n+1)]==f+1/2-(n+1)/2[cosn]+n/2[cos(n+1)]==f=[1-(n+1)cosn+ncos(n+1)]/[2(cos1-1)]。。