## College Algebra (11th Edition)

$-5,050$
$\bf{\text{Solution Outline:}}$ To evaluate the given expression, $\displaystyle\sum_{k=1}^{100} -k ,$ use the summation properties and then simplify. $\bf{\text{Solution Details:}}$ Using a property of the summation which is given by $\displaystyle\sum_{i=1}^n ci=c\displaystyle\sum_{i=1}^n i,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \displaystyle\sum_{k=1}^{100} (-1)k \\\\= -1\displaystyle\sum_{k=1}^{100} k \\\\= -\displaystyle\sum_{k=1}^{100} k .\end{array} Using a property of the summation which is given by $\displaystyle\sum_{i=1}^n i=\dfrac{n(n+1)}{2},$ with $n= 100 ,$ the expression above is equivalent to \begin{array}{l}\require{cancel} -\dfrac{n(n+1)}{2} \\\\= -\dfrac{100(100+1)}{2} \\\\= -5,050 .\end{array}