## College Algebra (11th Edition)

$-\dfrac{37}{20}$
$\bf{\text{Solution Outline:}}$ To evaluate the given summation expression, $\displaystyle\sum_{i=1}^5 \dfrac{x_i}{x_i+3} ,$ substitute $i$ with the values from $1$ to $5 .$ Then substitute the given values for each $x_i$'s $\bf{\text{Solution Details:}}$ Substituting $i$ with the numbers from $1$ to $5 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} \dfrac{x_1}{x_1+3}+\dfrac{x_2}{x_2+3}+\dfrac{x_3}{x_3+3}+\dfrac{x_4}{x_4+3}+\dfrac{x_5}{x_5+3} .\end{array} Using the given values $x_1=-2, x_2=-1, x_3=0, x_4=1,$ and $x_5=2,$ the expression above evaluates to \begin{array}{l}\require{cancel} \dfrac{-2}{-2+3}+\dfrac{-1}{-1+3}+\dfrac{0}{0+3}+\dfrac{1}{1+3}+\dfrac{2}{2+3} \\\\= \dfrac{-2}{1}+\dfrac{-1}{2}+\dfrac{0}{3}+\dfrac{1}{4}+\dfrac{2}{5} \\\\= -2-\dfrac{1}{2}+0+\dfrac{1}{4}+\dfrac{2}{5} \\\\= -2-\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{2}{5} \\\\= -\dfrac{40}{20}-\dfrac{10}{20}+\dfrac{5}{20}+\dfrac{8}{20} \\\\= -\dfrac{37}{20} .\end{array}