Let Tn be the nth term of the given series,
Then
Tn = (nth term of 1,5,9 …) × (nth term 3,7,11 …) × (nth term of 4,8,12 …)
= [1 + (n − 1)4] × [3 + (n − 1)4] × [4 + (n − 1)4]
= (4n − 3) × (4n − 1) × (4n)
= (16n2 − 16n + 3)4n
= 64n3 − 64n2 + 12n
Let Sn denote the sum to n term of the given series. Then,
Sn = \(\displaystyle\sum_{k=1}^{n} T_k = \displaystyle\sum_{k=1}^{n}(64n^3-64n^2+12n)\)
= \(64\displaystyle\sum_{k=1}^{n} n^3 -64\displaystyle\sum_{k=1}^{n}n^2+12\displaystyle\sum_{k=1}^{n}n\)
= 64 \(\big[\frac{n(n + 2)}{2} \big]^2\) − 64 \(\big[\frac{n(n + 1)(2n+1)}{6} \big]\) + 12 \(\big[\frac{n(n + 1)}{2} \big]\)
= \(\frac{64\{n(n + 1)\}^2}{4}\) − \(\frac{64\{n(n + 1)(2n+1)\}}{6}\) + 6n(n + 1)
= 16{n(n + 1)}2 − \(\frac{32\{n(n + 1)(2n+1)\}}{3}\) + 6n(n + 1)
= \(\frac{n(n + 1)}{3} \)[48{n(n + 1)} − 32(2n + 1) + 18]
= \(\frac{n(n + 1)}{3} \)[48(n2 + n) − (64n + 32) + 18]
= \(\frac{n(n + 1)}{3} \)[48n2 + 48n − 64n − 32 + 18]
= \(\frac{n(n + 1)}{3} \) [48n2 − 16n − 14]