The correct answer is (A):
N th term of the series can be written as
tn (4n 3)(4n 7)
= 16n2 + 40n + 21
∑tn = 16∑n2 + 40∑n + 21∑1
= 16n(n+1)(2n+1) / 6+40n(n+1) / 2+21
here n = 23 (7, 11, 15….. 95 is an AP with common different 4 with 23 terms)
∑tn = 16×23×24×47 / 6+20×23×24+21×23
= 80707