1st factors are 1, 4, 7,
∴ nthterm = 1 + (n – 1)3 = 3n – 2
2nd factor are 2, 5, 8,
∴ nth term = 2 + (n – 1) 3 = 3n – 1
⇒ nth term of the series is (3n – 2)(3n – 1)
∴ The sum = Σ(3n – 2)(3n – 1)
= Σ(9x2 – 9x + 2) = 9Σn2 -9Σn + 2Σ1
= 9((n(n + 1)(2n + 1))/6 - 9((n(n + 1))/2 + 2n
= 1/2n(n + 1)[3(2n + 1) – 9] + 2n = n(3n2 – 1).
