The sum S_N is defined by S_N = ∑_{n=1}^N n³. Using the identity (n+1)⁶ – (n-1)⁶ = 6n⁵ + 5n³ + 3n, find S_N in terms of N. [You need not simplify your result.]

Mark scheme

Marking points from the official mark scheme (May/June 2010). The mark or mark type is shown in brackets.

1. [1] (a) Apply difference method to the identity (N+1)⁶ – (N-1)⁶ = 6N⁵ + 5N³ + 3N
2. [1] (a) Correct LHS for sum: (N+1)⁶ + N⁶ - 1 - 0⁶ = 6S_N + 5S_N³ + 3S_N
3. [1] (a) Correct RHS for sum: ∑ [(n+1)⁶ – (n-1)⁶] = (N+1)⁶ + N⁶ - 1 - 0⁶ = 6∑n⁵ + 5∑n³ + 3∑n. So, 6S_N + (5/4)N²(N+1)² + 3N(N+1)/16
4. [1] (a) Expression for S_N: S_N = (1/6)[(N+1)⁶ + N⁶ - 1 - (5/4)N²(N+1)² - 3N(N+1)/16]
5. [2] (i) For λ = 6, lim N⁻⁶S_N = 1/6 (B2)
6. [1] (ii) For λ > 6, lim N⁻λS_N = 0 (B1)