\begin{aligned} &\begin{aligned} & s=1+6+\frac{9\left(1^2+2^2+3^2\right)}{7}+ \\ & \frac{12\left(1^2+2^2+3^2+4^2\right)}{9}+\frac{15\left(1^2+2^2+3^2+4^2+5^5\right)}{11}+\ldots \\ & s=\frac{3 \cdot(1)^2}{3}+\frac{6 \cdot\left(1^2+2^2\right)}{5}+\frac{9 \cdot\left(1^2+2^2+3^2\right)}{7}+ \\ & \frac{12 \cdot\left(1^2+2^2+3^2+4^2\right)}{9}+\ldots \end{aligned}\\ &\mathrm{N}^{\text {th }} \text { term of the series }\\ &\begin{aligned} & =t_n=\frac{3 n \cdot\left(1^2+2^2+\ldots+n^2\right)}{(2 n+1)} \\ & t_n=\frac{3 n \cdot n(n+1)(2 n+1)}{6(2 n+1)}=\frac{n^3+n^2}{2} \\ & S_n=\Sigma t_n=\frac{1}{2}\left\{\left(\frac{n(n+1)}{2}\right)^2+\frac{n(n+1)(2 n+1)}{6}\right\} \\ & =\frac{n(n+1)}{4}\left(\frac{n(n+1)}{2}+\frac{2 n+1}{3}\right) \\ & \therefore S_{15}=\frac{15 \times 16}{4}\left\{\frac{15 \cdot 16}{2}+\frac{31}{3}\right\} \\ & \quad \quad=60 \times 120+60 \times \frac{31}{3} \\ & \quad=7200+620 \\ & \quad=7820 \end{aligned} \text { } \begin{aligned} & \\ & \\ & \therefore \\ & \end{aligned} \end{aligned}