## College Algebra (6th Edition)

$\displaystyle \sum_{k=1}^{14}(2k+4)$ (sample answer)
$a_{1}=6$ $a_{2}=8=a_{1}+2=6+2$ $a_{3}=10=a_{2}+2=6+2+2=6+2(2)$ $a_{4}=12=a_{3}+2=6+2(2)+2=6+3(2)$ $...$ 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 $(2k+4)$ $\displaystyle \sum_{k=1}^{14}(2k+4)$