## Precalculus (6th Edition) Blitzer

$0$
Here, we have $a_1=-b_1=-4$, $a_2=-b_2=-2$,$a_3=-b_3=0$,$a_4=-b_4=2$,$a_5=-b_5=4$ Thus, $=[2(-4)+4]+[2(-2)+2]+[2(0)+0]+[2(2)+(-2)]+[2(4)+(-4)]$ $=-4-2+0+2+4$ $=0$