## Algebra 2 (1st Edition)

Let n be an even number, say 4. We will build a counterexample using $(-1)^{odd}=-1,\qquad (-1)^{even}=+1$ Let $a_{i}=b_{i}=(-1)^{i}$ Then, the sum on the LHS is $\displaystyle \sum_{i=1}^{4}a_{i}b_{i}=\sum_{i=1}^{4}(-1)^{i}(-1)^{i}$ $=\displaystyle \sum_{i=1}^{4}[(-1)(-1)]^{i}$ $=\displaystyle \sum_{i=1}^{4}1^{i}$ $=1+1+1+1=4$ but, $\displaystyle \sum_{i=1}^{4}a_{i}=(-1)^{1}+(-1)^{2}+(-1)^{3}+(-1)^{4}$ $=-1+1-1+1$ $=0$ The product on the RHS is $0.$ So, for this example, the LHS=4, and the RHS=0. The statement is not true.