Using mathematical induction, the numbers $a_n$'s are defined by $a_0=1, a_{n+1}= 3n^2+n+a_n, \left(n\ge 0\right)$, then $a_n=$

Given,  $a_0=1,a_{n+1}=3n^2+n+a_n \left(n\ge 0\right)$

Putting, $n=0$,  

$a_{0+1}=3\times 0+0+a_0$

$a_1=a_0=1$

$n=1$,

$a_{1+1}=3\times 1^2+1+a_1$

$a_2=3+1+1=5$

$n=2$.

$a_{2+1}=3{\left(2\right)}^2+2+5$

$a_3=19$

$a_n: a_3$

$\left(1\right) n^3+n^2+1 \Rightarrow 3^3+3^2+1\Rightarrow 37$, (wrong option)

$\left(2\right) n^3-n^2+1\Rightarrow 3^3-3^2+1\Rightarrow 19$, (right option)

$\left(3\right) n^3-n^2\Rightarrow 3^3-3^2=18$(wrong option)

$\left(4\right) n^3+n^2\Rightarrow 3^3+3^2\Rightarrow 36$(wrong option)