$a_{n+1}=a_n + 4n$

$a_{n+1}-a_n = 4n$

This is a telescoping series...
$a_2-a_1=4(1)$
$a_3-a_2=4(2)$ $\vdots$ $a_{18}-a_{17}=4(17)$

Adding all these equations,
$a_{18}-a_1 = 4(1+2+3+\ldots+17)$

$a_{18} = a_1 + 4\left(\frac{17(17+1)}{2}\right) = 5 + 2\times 17\times 18 = 5+612 = \boxed{617}$

On close observation..the series telescopes and can be expressed as

$a_{n+1-i}=a_{1}+4\times\displaystyle \sum_{i=1}^{n-i}n$

Now substituting $n=17 i=0$

$a_{18}=a_{1}+4(\frac {17(17+1)}{2})$ $\Rightarrow \boxed{a_{18}=617}$