Given an−an−1=2
So the sequence is an A.P. with first term a1=1 and common difference d=2
i.e., an=a1+(n−1)d=2n−1
Also, an=bn−bn−1
We know 2n−1=n2−(n−1)2
So, bn=n2
Now, n=1∑15an⋅bn=n=1∑15(2n3−n2)=2n=1∑15n3−n=1∑15n2
=2(215⋅16)2−615⋅16⋅31
=27560