#### Answer

$S_{60}=3,660$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To find the sum of the first $ 60 $ positive even integers, use the formula for finding the sum of $n$ terms that form an arithmetic sequence.
$\bf{\text{Solution Details:}}$
The sequence described by "the first $60$ positive even integers," is an arithmetic sequence with $a_1=2, d=2,$ and $n=60.$ Using the formula for the sum of the first $n$ terms of an airthmetic sequence, which is given by $ S_n=\dfrac{n}{2}[2a_1+(n-1)d] ,$ then \begin{array}{l}\require{cancel} S_{60}=\dfrac{60}{2}[2(2)+(60-1)2] \\\\ S_{60}=30[4+59(2)] \\\\ S_{60}=30[4+118] \\\\
S_{60}=30[122]
\\\\
S_{60}=3,660
.\end{array}