Answer: $\mathbf A$
Explanation:
The above series forms an $\mathbf {AP}$
where, $\mathrm {a = 3, \;d = 8,\;a_n = (8n-5)}$
So, sum of n terms of $\mathbf {AP}$ is given by:
$\begin {align} \mathrm {S_n} &= \mathrm {\frac{n}{2}[a + a_n] }\\&= \mathrm {\frac{n}{2}[3 + 8n-5] }\\&=\mathrm { \frac{n}{2}[8n-2] }\\&= \mathrm {n(4n-1)}\end {align}$
$\therefore \mathrm{S_n = 4n^2-n}$
$\therefore \mathbf A$ is the right option.