Let a 1 = b 1 = 1 , a n = a n - 1 + 2 and b n = a n + b n - 1 for every natural number n ≥ 2 . Then…

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, $\sum _{n=1}^{15}{a}_n·b_n=\sum _{n=1}^{15}\left(2n^3-n^2\right)=\underset{n=1}{\overset{15}{2\sum }}{n}^3-\sum _{n=1}^{15}{n}^2$

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

$=27560$