A2 June 2025 Paper 1 Q10
10 Astrid is solving this mathematics problem:
The series \(S_n\) is defined by \[S_n = 2 + 4 + 6 + \ldots + 2n \quad (n \in \mathbb{Z}, n \geqslant 1)\]Prove by induction that \[S_n = n(n + 1)\] |
Astrid’s solution is as follows:
Assume the result is true for \(n = k\) Then \[\begin{aligned} &S_k = k(k + 1) \\ &S_{k+1} = S_k + 2(k + 1) \\ &S_{k+1} = k(k + 1) + 2(k + 1) \\ &S_{k+1} = (k + 2)(k + 1) \\ &S_{k+1} = (k + 1)((k + 1) + 1) \end{aligned}\]So the result is also true for \(n = k + 1\) The result is true for \(n = 1\) It is true for \(n = k\), and also true for \(n = k + 1\) Hence, by induction \(S_n = n(n + 1)\) for all integers \(n \geqslant 1\) |
(a)
(i) Chloe says that Astrid missed out an essential part of the proof, which could have been written at the start.
Explain what Astrid missed out. [1 mark]
(ii) Write down the working that Astrid missed out. [1 mark]
(b)
(i) One statement in the last three lines of Astrid’s solution is written incorrectly.
Which statement is written incorrectly?
Tick (✓) one box. [1 mark]
- The result is true for \(n = 1\)
- It is true for \(n = k\), and also true for \(n = k + 1\)
- Hence, by induction \(S_n = n(n + 1)\) for all integers \(n \geqslant 1\)
(ii) Write out a correct statement which should replace the incorrect statement identified in part (b)(i) [1 mark]
| Scheme | Marks | AO |
|---|---|---|
| (i) States that Astrid did not show that the result is true for \(n = 1\) | E1 | 2.3 |
| (1) | ||
| (ii) Shows that the result is true for \(n = 1\) for both definition and formula | E1 | 2.4 |
| (1) |
Typical solution
(i)
Astrid did not show that the result is true for \(n = 1\)
(ii)
\[S_1 = 2\]and
\[S_1 = 1 \times (1 + 1) = 2\]| Scheme | Marks | AO |
|---|---|---|
| (i) Ticks 2nd answer. | E1 | 2.3 |
| (1) | ||
| (ii) Writes a correct statement to replace the incorrect statement. | R1 | 2.1 |
| (1) | ||
| (4 marks) |
Typical solution
(i)
It is true for \(n = k\), and also true for \(n = k + 1\)
(ii)
If true for \(n = k\), it is also true for \(n = k + 1\)