Jeet's answer is good enough, but if you are looking for another approach, it could be :
3 + 11 + ... + (8n-5) = $\sum (8n-5) = 8n(n+1)/2 - 5n = 4n^2 - n$
The summation function is $n(n+1)/2$ instead of $n(n-1)/2$ because in summation, the starting value of n is taken as 0, whereas, we want it to start from 1 here.