Let ${\mathrm{a}}_1={\mathrm{b}}_1=1, {\mathrm{a}}_{\mathrm{n}}={\mathrm{a}}_{\mathrm{n}-1}+2\text{ and }{\mathrm{b}}_{\mathrm{n}}={\mathrm{a}}_{\mathrm{n}}+{\mathrm{b}}_{\mathrm{n}-1}$ for every natural number $\mathrm{n}\ge 2$. Then $\sum _{\mathrm{n}=1}^{15} {\mathrm{a}}_{\mathrm{n}}\cdot {\mathrm{b}}_{\mathrm{n}}$ is equal to _______.

$\begin{array}{l}{\mathrm{a}}_1={\mathrm{b}}_1=1\\ {\mathrm{a}}_2={\mathrm{a}}_1+2=3,{\mathrm{a}}_3={\mathrm{a}}_2+2=5,{\mathrm{a}}_4={\mathrm{a}}_2+2=7\\ \Rightarrow \boxed{{\mathrm{a}}_{\mathrm{n}}=2\mathrm{n}-1}\\ {\mathrm{b}}_2={\mathrm{a}}_1+{\mathrm{b}}_1=4,{\mathrm{b}}_3={\mathrm{a}}_3+{\mathrm{b}}_2=9\\ {\mathrm{b}}_4={\mathrm{a}}_4+{\mathrm{b}}_3=16.\\ {\mathrm{b}}_{\mathrm{n}}={\mathrm{n}}^2\\ \\ \sum _{\mathrm{n}=1}^{15} {\mathrm{a}}_{\mathrm{n}}{\mathrm{b}}_{\mathrm{n}}=\sum _{\mathrm{n}=1}^{15} (2\mathrm{n}-1){\mathrm{n}}^2\\ \end{array}$

$\begin{array}{l}\sum _{\mathrm{n}=1}^{15} \left(2{\mathrm{n}}^3-{\mathrm{n}}^2\right)\\ =2\frac{{\mathrm{n}}^2(\mathrm{n}+1{)}^2}{4}-\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}\end{array}$

Put n = 15

$\begin{array}{l}=\frac{2\times 225\times 16\times 16}{4}-\frac{15\times 16\times 31}{6}\\ =27560\end{array}$