Find the sum of the following series up to n terms:

(1³ / 1) + ((1³ + 2³) / (1 + 3)) + ((1³ + 2³ + 3³) / (1 + 3 + 5)) + …

The nᵗʰ term of the given series is:

Tₙ = (1³ + 2³ + 3³ + … + n³) / (1 + 3 + 5 + … + (2n − 1))

We know that:

1³ + 2³ + 3³ + … + n³ = [n(n + 1)/2]²

and 1 + 3 + 5 + … + (2n − 1) = n²

∴ Tₙ = [n²(n + 1)² / 4n²] = [(n + 1)² / 4] = (1/4)n² + (1/2)n + (1/4)

Sum of n terms:

Sₙ = ΣTₙ = Σ[(1/4)n² + (1/2)n + (1/4)]

= (1/4)Σn² + (1/2)Σn + (1/4)Σ1

= (1/4)[n(n + 1)(2n + 1)/6] + (1/2)[n(n + 1)/2] + (1/4)n

= [n((n + 1)(2n + 1) + 6(n + 1) + 6)] / 24

= [n(2n² + 9n + 13)] / 24

Hence, the sum of the series up to n terms is:

Sₙ = (n(2n² + 9n + 13)) / 24