Expand and simplify (r + 1)⁴ – r⁴. Use the method of differences together with the standard results for Σr and Σr² to show that Σ(from r=1 to n) r³ = ¼n²(n + 1)².
- Expand and simplify (r + 1)⁴ – r⁴.[1]
- Use the method of differences together with the standard results for Σr and Σr² to show that Σ(from r=1 to n) r³ = ¼n²(n + 1)².[4]
Mark scheme
Marking points from the official mark scheme (May/June 2014). The mark or mark type is shown in brackets.
- [B1] (r + 1)⁴ – r⁴ = 4r³ + 6r² + 4r + 1
- [M1] (Σ(from r=1 to n) (r+1)⁴ – r⁴) = 4Σ(from r=1 to n) r³ + 6Σ(from r=1 to n) r² + 4Σ(from r=1 to n) r + Σ(from r=1 to n) 1
- [A1] (n + 1)⁴ – 1⁴ = 4Σ(from r=1 to n) r³ + 6Σ(from r=1 to n) r² + 4Σ(from r=1 to n) r + n
- [A1] n⁴ + 4n³ + 6n² + 4n + 1 – 1 = 4Σ(from r=1 to n) r³ + n(n+1)(2n+1) + 2n(n+1) + n
- [A1] ⇒ ... Σ(from r=1 to n) r³ = ¼n²(n + 1)². (AG)
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