Answer
$\sum_{i=1}^{14}(2i+4)$
Work Step by Step
Since, we have $6+8+10+12+.....32$
This form an arithmetic progression sequence whose first term is $6$ and last term is $32$ with common difference of $2$.
For an arithmetic progression sequence, we have
$a_n=a_1+(n-1)d$
thus, $a_n=6+(n-1)2=6+2n-2=2n+4$
when $n=14$
Lat term becomes $a_{14}=2(14)+4=32$
Hence, $6+8+10+12+.....32$ can be summed up with the summation notation as:
$\sum_{i=1}^{14}(2i+4)$
