AS June 2025 Paper 1 Q6
6. Prove by induction that for all positive integers \(n\)
\[\sum_{r=1}^{n} r^3 = \frac{1}{4}n^2(n+1)^2\](6)
| Scheme | Marks | AO |
|---|---|---|
| \(n = 1\) LHS = 1 RHS \(= \dfrac{1}{4}(1)^2(2)^2 = 1\) | B1 | 2.2a |
| Assume true for \(\boldsymbol{n = k}\) or \(\displaystyle\sum_{r=1}^{k} r^3 = \frac{1}{4}k^2(k+1)^2\) | M1 | 2.5 |
| \(\displaystyle\sum_{r=1}^{k+1} r^3 = \frac{1}{4}k^2(k+1)^2 + (k+1)^3\) Or \(\displaystyle(k+1)^3 = \sum_{r=1}^{k+1} r^3 - \frac{1}{4}k^2(k+1)^2\) or \((k+1)^3 = \dfrac{1}{4}(k+1)^2(k+2)^2 - \dfrac{1}{4}k^2(k+1)^2\) | M1 | 2.1 |
| \(\dfrac{1}{4}(k+1)^2\left[k^2 + 4(k+1)\right]\) Or Multiples out \(\dfrac{1}{4}k^2(k+1)^2 + (k+1)^3\) and \(\dfrac{1}{4}(k+1)^2(k+2)^2\) to form a quartic for each Or Multiplies out or factorises out \(\dfrac{1}{4}(k+1)^2(k+2)^2 - \dfrac{1}{4}k^2(k+1)^2\) | M1 | 1.1b |
| \(\dfrac{1}{4}(k+1)^2\left[k^2 + 4k + 4\right] = \dfrac{1}{4}(k+1)^2(k+2)^2\) leading to \(\dfrac{1}{4}(k+1)^2\left(\{k+1\} + 1\right)^2\) Or \(k^4 + 6k^3 + 13k^2 + 12k + 4 = k^4 + 6k^3 + 13k^2 + 12k + 4\) And draws conclusion true Or Shows \(\dfrac{1}{4}(k+1)^2(k+2)^2 - \dfrac{1}{4}k^2(k+1)^2 = (k+1)^3\) by either multiplying out both sides to reach the same cubic, and draws a conclusion or by factorisation | A1 | 1.1b |
| If true for \(n = k\) then true for \(n = k + 1\), and as also true for \(n = 1\), so the result is true for all \(n\) | A1 | 2.4 |
| (6) | ||
| (6 marks) |
Notes
(Corrected from the printed mark scheme: in the alternative lines the scheme prints \(\dfrac{1}{4}(k+1)^2(k+1)^2\) in place of \(\dfrac{1}{4}(k+1)^2(k+2)^2\), here and in the notes below; \(\sum_{r=1}^{k+1} r^3 = \dfrac{1}{4}(k+1)^2(k+2)^2\).)
B1: Shows the statement is true for \(n = 1\). Minimum requirement \(1 = \dfrac{1}{4}(1)(4)\)
M1: Makes the inductive assumption, assume true \(\boldsymbol{n = k}\). This may appear in the conclusion.
M1: A correct statement for \(n = k + 1\)
M1: Factorises out \((k+1)^2\) or multiples out \(\dfrac{1}{4}k^2(k+1)^2 + (k+1)^3\) and \(\dfrac{1}{4}(k+1)^2(k+2)^2\) to form a quartic for each
A1: Achieves \(\dfrac{1}{4}(k+1)^2\left(\{k+1\} + 1\right)^2\) with no omissions or errors. Shows both are sides are \(k^4 + 6k^3 + 13k^2 + 12k + 4\) no omissions or errors. and draws a minimal conclusion. Shows \(\dfrac{1}{4}(k+1)^2(k+2)^2 - \dfrac{1}{4}k^2(k+1)^2 = (k+1)^3\) may expand both sides and draw a minimal conclusion or factorises, no omissions or errors.
Note: they may state aiming for \(\dfrac{1}{4}(k+1)^2(k+2)^2\) and then achieves this which is fine for A1
A1: Makes appropriate concluding sentence covering the points indicated in scheme. Dependent on all previous marks except B1. The statement true for \(n = 1\) could appear at the start of the solution