June 2025 Paper 2 Q5
5 Prove that the sum of the first \(n\) positive odd numbers is a square number. [3]
| Scheme | Marks | AO |
|---|---|---|
| \(1 + 3 + 5 + \ldots + (2n \pm 1)\) | M1 | 2.1 |
| \(\dfrac{n}{2}(2\times 1 + (n-1)\times 2)\) or \(\dfrac{n}{2}(1 + (2n-1))\) | A1 | 1.1 |
| \(\dfrac{n}{2}\times 2n = n^2\) [which is a square number] | A1 | 2.2a |
| [3] |
Notes
M1: or identify AP with \(n\) terms and \(a = 1\), \(d = 2\); may be implied by award of first A1
A1: use of formula for sum of arithmetic progression with \(n\) terms and \(a = 1\) and \(l = 2n - 1\); or \(a = 1\) and \(d = 2\); allow \(\frac{n}{2}(2 + 2n - 2)\) but no further contraction
A1: correct intermediate step needed for A1
Alternative method (Further Maths 1)
| Scheme | Marks |
|---|---|
| \(2\sum_{r=1}^{n} r - n\) | M1 |
| \(= 2\times\dfrac{n(n+1)}{2} - n\) | A1 |
| \(n^2 + n - n = n^2\) [which is a square number] | A1 |
M1: from \(\sum_{r=1}^{n}(2r-1)\); allow for \(\sum_{r=1}^{n}(2r+1)\); may be implied by award of first A1
A1: all correct;
A1: correct intermediate step needed for A1
Alternative method (Further Maths 2)
| Scheme | Marks |
|---|---|
| assume \(\sum_{r=1}^{k}(2r-1) = k^2\) (i) working towards \(\sum_{r=1}^{k+1}(2r-1) = (k+1)^2\) (ii) | M1 |
| \(\sum_{r=1}^{k}(2r-1) + (2(k+1) - 1)\) \(= k^2 + 2k + 1\) \(= (k+1)^2\) so true for \(n = k + 1\) if true for \(n = k\) | A1 |
| \(2\times 1 - 1 = 1^2\) so true for \(n = 1\) so by induction true for \(n = 2, 3, 4, \ldots\) | A1 |
A1: correct intermediate step and statement needed