#### Answer

$\displaystyle \sum_{k=1}^{14}(2k+4)$
(sample answer)

#### Work Step by Step

$a_{1}=6$
$a_{2}=8=a_{1}+2=6+2$
$a_{3}=10=a_{2}+2=6+2+2=6+2(2)$
$a_{4}=12=a_{3}+2=6+2(2)+2=6+3(2)$
$...$
The pattern suggests that $a_{k}=6+(k-1)2$.
$a_{k}=6+(k-1)2$
$a_{k}=6+2k-2$
$a_{k}=2k+4$
The first (for k=1) term is $6$, the last term is $32$.
Find the last k:
$2k+4=32\quad/-4$
$2k=28\quad/\div 2$
$k=14$
So, we have a sum for k=1 to $14$,
and the general term is $(2k+4)$
$\displaystyle \sum_{k=1}^{14}(2k+4)$