Let a _ 1 = b _ 1 = 1, a _ n = a _ n - 1 + 2 and b _ n = a _ n + b...

Given ${a}_{n} - {a}_{n - 1} = 2$

So the sequence is an A.P. with first term ${a}_{1} = 1$ and common difference $d = 2$

i.e., ${a}_{n} = {a}_{1} + \left(n - 1\right)d = 2n - 1$

Also, ${a}_{n} = {b}_{n} - {b}_{n - 1}$

We know $2n - 1 = {n}^{2} - {\left(n - 1\right)}^{2}$

So, ${b}_{n} = {n}^{2}$

Now, $\mathop{∑}_{n = 1}^{15}{a}_{n} \cdot {b}_{n} = \mathop{∑}_{n = 1}^{15}\left(2{n}^{3} - {n}^{2}\right) = \mathop{2 ∑}_{n = 1}^{15}{n}^{3} - \mathop{∑}_{n = 1}^{15}{n}^{2}$

$= 2{\left(\dfrac{15 \cdot 16}{2}\right)}^{2} - \dfrac{15 \cdot 16 \cdot 31}{6}$

$= 27560$