Find the sum to n terms: 1² + (1² + 2²) + (1² + 2² + 3²) + …

The given series is 1² + (1² + 2²) + (1² + 2² + 3²) + … + aₙ, where aₙ = (1² + 2² + 3² + … + n²)

aₙ = n(n + 1)(2n + 1) / 6

= n(2n² + 3n + 1) / 6 = (2n³ + 3n² + n) / 6

= (1/3)n³ + (1/2)n² + (1/6)n

∴ Sₙ = Σaₙ = Σ[(1/3)k³ + (1/2)k² + (1/6)k]

= (1/3)Σk³ + (1/2)Σk² + (1/6)Σk

= (1/3)[n²(n + 1)² / 4] + (1/2)[n(n + 1)(2n + 1) / 6] + (1/6)[n(n + 1) / 2]

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

= (n(n + 1)/6)[(n² + n + 2n + 2)/2]

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

∴ Sₙ = n(n + 1)²(n + 2) / 12