Mathematics - Sequence & Series Question with Solution | TestHub

$t_n=\frac{3n\left(1^2+2^2+3^2+\dots +n^2\right)}{2n+1}=\frac{3n\times n\left(n+1\right)\left(2n+1\right)}{6 \left(2n+1\right)}$

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

$\therefore S_{15}=\sum _{n=1}^{15}{t}_n=\sum _{n=1}^{15}\frac{1}{2}\left(n^3+n^2\right)$

$=\frac{1}{2}\left[\sum _{n=1}^{15}{n}^3+\sum _{n=1}^{15}{n}^2\right]$

$=\frac{1}{2}\times \left[{\left(\frac{15\times 16}{2}\right)}^2+\frac{15\times 16\times 31}{6} \right]$   $\left\{\sum n^3={\left(\frac{n\left(n+1\right)}{2}\right)}^2, \sum n^2=\left(\frac{n\left(n+1\right)\left(n+2\right)}{6}\right)\right\}$

$=7200+620$

$=7820$