Answer
See below.
Work Step by Step
$S_k:2+7+12+...+(5k-3)=\frac{k(5k-1)}{2}$
$S_{k+1}:2+7+12+...+(5(k+1)-3)=\frac{(k+1)(5(k+1)-1)}{2}\\S_{k+1}:2+7+12+...+(5k+2)=\frac{(k+1)(5k+4)}{2}$
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