AS June 2018 Paper 1 Q10
10
(a) Prove by induction that, for all integers \(n \geqslant 1\),\[\sum_{r=1}^{n} r^3 = \frac{1}{4}n^2(n + 1)^2\]
[4 marks]
(b) Hence show that\[\sum_{r=1}^{2n} r(r - 1)(r + 1) = n(n + 1)(2n - 1)(2n + 1)\]
[4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Demonstrates the rule is correct for \(n = 1\) | B1 | 1.1b |
| States the rule is true for \(n = k\) and adds \((k + 1)^3\) to \(\frac{1}{4}k^2(k + 1)^2\) | M1 | 2.4 |
| Obtains \(\frac{1}{4}(k + 1)^2(k + 2)^2\) from \(\frac{1}{4}k^2(k + 1)^2 + (k + 1)^3\) | A1 | 2.2a |
| Completes a rigorous argument and explains how their argument proves the required result, This mark is only available if all previous marks have been awarded. | R1 | 2.1 |
Typical solution
\[\sum_{r=1}^{1} r^3 = 1^3 = 1 \quad \text{and} \quad \frac{1}{4} \times 1^2 \times 2^2 = 1\]\(\therefore\) it is true for \(n = 1\)
Assume it is true for \(n = k\)
\[\sum_{r=1}^{k} r^3 = \frac{1}{4}k^2(k + 1)^2\]\[\sum_{r=1}^{k} r^3 + (k + 1)^3 = \frac{1}{4}k^2(k + 1)^2 + (k + 1)^3\]\[\sum_{r=1}^{k+1} r^3 = \frac{1}{4}(k + 1)^2\left(k^2 + 4(k + 1)\right)\]\[\sum_{r=1}^{k+1} r^3 = \frac{1}{4}(k + 1)^2(k^2 + 4k + 4)\]\[\sum_{r=1}^{k+1} r^3 = \frac{1}{4}(k + 1)^2(k + 2)^2\]\(\therefore\) it is also true for \(n = k + 1\)
True for \(n = 1\), and true for \(n = k \Rightarrow\) true for \(n = k + 1\), then by induction it is true for all integers \(n \geqslant 1\)
AG
| Scheme | Marks | AO |
|---|---|---|
| Expresses LHS as summations of \(r^3\) and \(r\) Ignore limits of the sums in this part only. | B1 | 1.1b |
| Expresses LHS in terms of \(n\), using part (a) and \(\sum r = \frac{1}{2}n(n + 1)\) | M1 | 1.1a |
| Takes out \(n(2n + 1)\) as a factor or obtains \(4n^4 + 4n^3 - n^2 - n\) Allow one slip in second bracket or one incorrect term in the expansion. | M1 | 1.1a |
| Completes fully correct proof to reach the required result. This mark is only available if all previous marks have been awarded. Note: \(n(n + 1)(2n - 1)(2n + 1) = 4n^4 + 4n^3 - n^2 - n\) | R1 | 2.1 |
| (8 marks) |
Typical solution
\[\begin{aligned}\sum_{r=1}^{2n} r(r - 1)(r + 1) &= \sum_{r=1}^{2n}(r^3 - r) \\ &= \sum_{r=1}^{2n} r^3 - \sum_{r=1}^{2n} r \\ &= \frac{1}{4}(2n)^2(2n + 1)^2 - \frac{1}{2}2n(2n + 1) \\ &= n^2(2n + 1)^2 - n(2n + 1) \\ &= n(2n + 1)\left(n(2n + 1) - 1\right) \\ &= n(2n + 1)(2n^2 + n - 1) \\ &= n(2n + 1)(2n - 1)(n + 1) \\ &= n(n + 1)(2n - 1)(2n + 1)\end{aligned}\]AG