The sum of the following series 1+6+(9(1^2+2^2+3^2))/7+(12(1^2+2^2+3^2+4^2))/9; +(15(1^2+2^2+…+5^2))/11+⋯ up to 15 terms, is:

The sum of the following series $$\begin{aligned} & 1+6+\frac{9\left(1^2+2^2+3^2\right)}{7}+\frac{12\left(1^2+2^2+3^2+4^2\right)}{9} \\ & +\frac{15\left(1^2+2^2+\ldots+5^2\right)}{11}+\cdots \text { up to } 15 \text { terms, } \end{aligned} $$ is :