Question

Find the sum of the series (3³ – 2³) + (5³ – 4³) + (7³ – 6³) + … to n terms

Answer 1

Given series

⇒ (3³ – 2³) + (5³ – 4³) + (7³ – 6³) + ...

= (3³ + 5³ + 7³ + …) – (2³ + 4³ + 6³ + …)

⇒ [3³ + 5³ + 7³ + … (2n + 1)³] – [2³ + 4³ + 6³ + … (2n)³] 

∴ T_n = (2n + 1)³ – (2n)³

= (2n + 1 – 2n) [2n + 1)² + (2n + 1)(2n) + (2n)²]  ....[∵ a³ – b³ = (a – b)(a² + ab + b²)]

= 1 · [4n² + 1 + 4n + 4n² + 2n + 4n²]

= 12n² + 6n + 1

S_n = `sum "T"_n = 12 sum n^2 + 6 sum n + n`

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

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

= n[2(n + 1)(2n + 1) + 3(n + 1) + 1]

= n[2(2n² + 3n + 1) + 3n + 3 + 1]

= n[4n² + 6n + 2 + 3n + 4]

= n[4n² + 9n + 6]

= 4n³ + 9n² + 6n