![{(a - b)^2} + ({a^2} + {b^2}) + {(a + b)^2} + s.... + [{(a + b)^2} + 6ab]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8a1c84de18d7934d0b7f67915b7f67db_l3.png "Rendered by QuickLaTeX.com")
Answer:
First term, a = a1 = (a-b)2
Common difference, d = a2 – a1 = (a2 + b2) – (a – b)2 = 2ab
an term of given AP is [(a + b)2 + 6ab]
an = a + (n-1) d
[(a + b)2 + 6ab] = (a-b)2 + (n-1)2ab
a2 + b2 + 2ab + 6ab = a2 + b2 – 2ab + 2abn – 2ab
a2 + b2 + 8ab – a2 – b2 + 2ab + 2ab = 2abn
12ab = 2abn
n = 12ab / 2ab
n = 6
S = n/2 (a + l)
S = 6/2 ((a-b)2 + [(a + b)2 + 6ab])
S = 3 (a2 + b2 – 2ab + a2 + b2 + 2ab + 6ab)
S = 3 (2a2 + 2b2 + 6ab)
S = 3 × 2 (a2 + b2 + 3ab)
S = 6 (a2 + b2 + 3ab)
∴ The sum of the series is 6 (a2 + b2 + 3ab)