## College Algebra 7th Edition

We evaluate the sum: $\sum_{j=1}^{100}(-1)^{j}=(-1)^{1}+(-1)^{2}+(-1)^{3}+(-1)^{4}+(-1)^{5}+...+(-1)^{99}+(-1)^{100}=-1+1-1+1-1+...-1+1=0$ (Since we have an even number of terms, the $+1$ and $-1$ cancel out.)