Precalculus (6th Edition) Blitzer

False. The correct expression is $\sum\limits_{i=1}^{2}{{{a}_{i}}{{b}_{i}}\ne \sum\limits_{i=1}^{2}{{{a}_{i}}}}\sum\limits_{i=1}^{2}{{{b}_{i}}}$
The given expression is $\sum\limits_{i=1}^{2}{{{a}_{i}}{{b}_{i}}=\sum\limits_{i=1}^{2}{{{a}_{i}}}}\sum\limits_{i=1}^{2}{{{b}_{i}}}$ For the statement to be true, the left side $\sum\limits_{i=1}^{2}{{{a}_{i}}{{b}_{i}}}$ has to be equal to the right side, $\sum\limits_{i=1}^{2}{{{a}_{i}}}\sum\limits_{i=1}^{2}{{{b}_{i}}}$ For the right hand side: $\sum\limits_{i=1}^{2}{{{a}_{i}}{{b}_{i}}}$ $={{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}$ For the left hand side: $\sum\limits_{i=1}^{2}{{{a}_{i}}}\sum\limits_{i=1}^{2}{{{b}_{i}}}$. Hence, the left hand side is not equal to the right hand side.