The sum of the following series 1+6+912+22+327+1212+22+32+429+1512+22+…+5211+… upto 15 terms, is

The $n^{th}$ term of given expression is

$T_{n} = \frac{(3 + (n - 1) 3) (1^{2} + 2^{2} + 3^{2} + \dots + n^{2})}{(2 n + 1)}$

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

Now, $S_{15} = \frac{1}{2} \sum_{n = 1}^{15} (n^{3} + n^{2})$ $= \frac{1}{2} [{(\frac{15 (15 + 1)}{2})}^{2} + \frac{15 \times 16 \times 31}{6}]$

$= \frac{1}{2} [(120)^{2} + 1240] = 7820$

The sum of the following series $1 + 6 + \frac{9 (1^{2} + 2^{2} + 3^{2})}{7} + \frac{12 (1^{2} + 2^{2} + 3^{2} + 4^{2})}{9}$ $+ \frac{15 (1^{2} + 2^{2} + \dots + 5^{2})}{11} + \dots$ upto 15 terms, is