Question

Find the sum of the series (2³ – 1³) + (4³ – 3³) + (6³ – 15³) + ... to 

1. n terms
2. 8 terms

Answer 1

(2³ – 1³) + (4³ – 3³) + (6³ – 15³) + ...

The n^th term is

T_n ​= (2n)³ − (2n − 1)³

Tn​​ = 8n³ − (2n − 1)³

= 8n³ − (8n³ − 12n² + 6n − 1)

= 12n² − 6n + 1

`S_n = sum_(k=1)^n (12k^2 - 6k + 1)`

`sum_(k=1)^n k = (n(n+1))/2, sum_(k=1)^n k^2 = (n(n+1)(2n+1))/6`

`S_n = 12 * (n(n+1)(2n+1))/6 - 6 * (n(n+1))/2 + n`

= 2n(n + 1) (2n + 1) − 3n(n + 1) + n

= n[2(n + 1)(2n + 1) − 3(n + 1) + 1]

= n(4n² + 3n)

= n² (4n + 3)

Substitute n = 8:

S₈ ​= 8²(4 × 8 + 3)

= 64(35)

= 2240​