Answer
$\displaystyle \sum_{k=1}^{14}(2k+3)$
Work Step by Step
$a_{1}=5$
$a_{2}=7=a_{1}+2=5+2$
$a_{3}=9=a_{2}+2=5+2+2=5+2(2)$
$a_{4}=11=a_{3}+2=5+2(2)+2=5+3(2)$
$...$
The pattern suggests that $a_{k}=5+(k-1)2$.
$a_{k}=5+(k-1)2$
$a_{k}=5+2k-2$
$a_{k}=2k+3$
The first (for k=1) term is 5, the last term is 31.
Find the last k:
$2k+3=31\quad/-3$
$2k=28\quad/\div 2$
$k=14$
So, we have a sum for k=1 to $14$,
and the general term is $(2k+3)$
$\displaystyle \sum_{k=1}^{14}(2k+3)$
