Question

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

Answer 1

(2³ – 1³) + (4³ – 3³) + (6³ – 15³) + ... to n terms

`sum_1^"n" 2^3 + 4^3 + 6^3 + ... - sum_1^"n" (1^3 + 3^3 + 5^3 + ...)`

`sum_1^"n" [(2"n")^3 - (2"n" - 1)^3]`

a³ – b³ = (a – b)³ + 3ab(a – b)

= (2n – 2n + 1)³ + 3(2n)(2n – 1)(2n – 2n + 1)

= 1³ – 3.(2n) (2n – 1)

= 1 + 6n(2n – 1)

= 1 + 12n² – 6n

= `sum(1 + 12"n"^2 - 6"n")`

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

= n + 2n (n + 1)(2n + 1) – 3n(n + 1)

= n + (2n² + 2n) (2n + 1) – 3n² – 3n

= n + 4n³ + 4n² + 2n² + 2n – 3n² – 3n

 = 4n³ + 3n² = sum of ‘n’ terms.