A2 June 2022 Paper 2 Q5
5 Prove by induction that, for all integers \(n \geqslant 1\),
\[\sum_{r=1}^{n} r^3 = \left\{\frac{1}{2}n(n + 1)\right\}^2\][4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Demonstrates the result for \(n = 1\) and states that it is true for \(n = 1\) | B1 | 2.2a |
| Assumes the result true for \(n = k\) (PI by algebraic working) and adds \((k + 1)^3\) to \(= \left(\dfrac{1}{2}k(k + 1)\right)^2\) | M1 | 2.1 |
| Obtains \(= \dfrac{1}{4}(k + 1)^2(k + 2)^2\) from \(= \left(\dfrac{1}{2}k(k + 1)\right)^2 + (k + 1)^3\) | A1 | 1.1b |
| Concludes a reasoned argument by stating that \(\displaystyle\sum_{r=1}^{n} r^3 = \left\{\frac{1}{2}n(n + 1)\right\}^2\) is true for \(n = 1\), and that \(\displaystyle\sum_{r=1}^{k} r^3 = \left(\frac{1}{2}k(k + 1)\right)^2\) implies \(\displaystyle\sum_{r=1}^{k+1} r^3 = \left(\frac{1}{2}(k + 1)(k + 2)\right)^2\) And hence, by induction, that \(\displaystyle\sum_{r=1}^{n} r^3 = \left\{\frac{1}{2}n(n + 1)\right\}^2\) is true for all integers \(n \geqslant 1\) | R1 | 2.1 |
| (4 marks) |
Typical solution
Let \(n = 1\)
Then \(\left(\dfrac{1}{2}n(n + 1)\right)^2 = 1^2 = 1\) and \(\displaystyle\sum_{r=1}^{n} r^3 = 1\)
So the result is true for \(n = 1\)
Assume the result is true for \(n = k\)
Then
\[\sum_{r=1}^{k} r^3 = \left(\frac{1}{2}k(k + 1)\right)^2\]and
\[\begin{aligned}\sum_{r=1}^{k+1} r^3 &= \left(\frac{1}{2}k(k + 1)\right)^2 + (k + 1)^3 \\ &= \frac{1}{4}(k + 1)^2\left(k^2 + 4(k + 1)\right) \\ &= \frac{1}{4}(k + 1)^2(k^2 + 4k + 4) \\ &= \frac{1}{4}(k + 1)^2(k + 2)^2 \\ &= \left(\frac{1}{2}(k + 1)(k + 2)\right)^2\end{aligned}\]The result is true for \(n = 1\); if true for \(n = k\), then it’s also true for \(n = k + 1\) and hence by induction
\[\sum_{r=1}^{n} r^3 = \left\{\frac{1}{2}n(n + 1)\right\}^2\]for all integers \(n \geqslant 1\)