Question

Find the sum of the series,

1. 3. 4 + 5. 7. 8 + 9. 11. 12 + ⋯ upto n terms.

Answer 1

Let T_n be the n^th term of the given series,

Then

T_n = (n^th term of 1,5,9 …) × (n^th term 3,7,11 …) × (n^th 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 S_n denote the sum to n term of the given series. Then,

S_n = \(\displaystyle\sum_{k=1}^{n} T_k = \displaystyle\sum_{k=1}^{n}(64n^3-64n^2+12n)\)

= \(64\displaystyle\sum_{k=1}^{n} n^3 -64\displaystyle\sum_{k=1}^{n}n^2+12\displaystyle\sum_{k=1}^{n}n\)

= 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]