Answer: (4)

Solution

$n^{th}$ term of the series,

$t_n = \frac{3n \cdot (1^2 + 2^2 + \ldots + n^2)}{(2n+1)}$

$\Rightarrow t_n = \frac{3n \cdot n(n+1)(2n+1)}{6(2n+1)} = \frac{n^2 + n^2}{2}$

$\therefore S_n = \sum t_n = \frac{1}{2}\left[\left(\frac{n(n+1)}{2}\right)^2 + \frac{n(n+1)(2n+1)}{6}\right]$

$= \frac{n(n+1)}{4}\left(\frac{n(n+1)}{2} + \frac{2n+1}{3}\right)$

Hence, sum of the series upto 15 terms is,

$S_{15} = \frac{15 \times 16}{4}\left[\frac{15.16}{2} + \frac{31}{3}\right]$

$= 60 \times 120 + 60 \times \frac{31}{3}$

$= 7200 + 620$

$= 7820$