AS June 2023 Paper 1 Q13
13
(a) Prove by induction that, for all integers \(n \geqslant 1\),\[\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n + 1)(2n + 1)\] [4 marks]
(b) Hence, or otherwise, write down a factorised expression for the sum of the first \(2n\) squares\[1^2 + 2^2 + 3^2 + \ldots + (2n)^2\] [1 mark]
(c) Use the formula in part (a) to write down a factorised expression for the sum of the first \(n\) even squares\[2^2 + 4^2 + 6^2 + \ldots + (2n)^2\] [2 marks]
(d) Hence, or otherwise, show that the sum of the first \(n\) odd squares is\[an(bn - 1)(bn + 1)\]
where \(a\) and \(b\) are rational numbers to be determined. [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Substitutes \(n = 1\) into LHS and RHS of \(\displaystyle\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n + 1)(2n + 1)\) | B1 | 2.2a |
| Uses \(\displaystyle\sum_{r=1}^{k} r^2 = \frac{1}{6}k(k + 1)(2k + 1)\) and considers \(\displaystyle\sum_{r=1}^{k} r^2 + (k + 1)^2\) Condone use of \(n\) in place of \(k\) | M1 | 2.4 |
| Completes working to show \(\dfrac{1}{6}k(k + 1)(2k + 1) + (k + 1)^2\) is equivalent to \(\dfrac{1}{6}(k + 1)(k + 1 + 1)(2(k + 1) + 1)\) Accept \(\dfrac{1}{6}(k + 1)(k + 2)(2k + 3)\) | A1 | 2.2a |
| Completes a reasoned argument by stating that the rule is true for \(n = 1\) and if the rule is true for \(n = k\) then it is also true for \(n = k + 1\) and concludes that by induction \(\displaystyle\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n + 1)(2n + 1)\) is true for all integers \(n \geqslant 1\) This mark is dependent on all previous marks. The algebra must use an alternative letter to \(n\) Condone reference to ‘rule’/ ‘statement’ / ‘it’ in the concluding statement. | R1 | 2.1 |
| (4) |
Typical solution
When \(n = 1\):
\[\sum_{r=1}^{1} r^2 = 1^2 = 1 \quad \text{and} \quad \frac{1}{6} \times 1 \times 2 \times 3 = 1\]\(\therefore\) the rule is true for \(n = 1\)
Assume it is true for \(n = k\)
\[\sum_{r=1}^{k} r^2 = \frac{1}{6}k(k + 1)(2k + 1)\]\[\Rightarrow \sum_{r=1}^{k} r^2 + (k + 1)^2 = \frac{1}{6}k(k + 1)(2k + 1) + (k + 1)^2\]\[\begin{aligned}\Rightarrow \sum_{r=1}^{k+1} r^2 &= \frac{1}{6}(k + 1)\big(k(2k + 1) + 6(k + 1)\big) \\ &= \frac{1}{6}(k + 1)(2k^2 + 7k + 6) \\ &= \frac{1}{6}(k + 1)(k + 2)(2k + 3) \\ &= \frac{1}{6}(k + 1)(k + 1 + 1)(2(k + 1) + 1)\end{aligned}\]\(\therefore\) the rule is also true for \(n = k + 1\)
So, by induction,
\[\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n + 1)(2n + 1)\]is true for all integers \(n \geqslant 1\)
| Scheme | Marks | AO |
|---|---|---|
| Writes the correct expression. Accept partial factorisation, eg \(\dfrac{n}{3}(8n^2 + 6n + 1)\) ISW | B1 | 1.1b |
| (1) |
Typical solution
\[\begin{aligned} &\frac{1}{6} \times 2n(2n + 1)(4n + 1) \\ &= \frac{1}{3}n(2n + 1)(4n + 1)\end{aligned}\]| Scheme | Marks | AO |
|---|---|---|
| Writes the required sum as a multiple of \(\displaystyle\sum_{r=1}^{n} r^2\) | M1 | 3.1a |
| Obtains the correct expression. Accept partial factorisation, eg \(\dfrac{2n}{3}(2n^2 + 3n + 1)\) ISW | A1 | 1.1b |
| (2) |
Typical solution
\[\begin{aligned}\sum_{r=1}^{n} (2r)^2 &= 4\sum_{r=1}^{n} r^2 \\ &= 4 \times \frac{1}{6}n(n + 1)(2n + 1) \\ &= \frac{2}{3}n(n + 1)(2n + 1)\end{aligned}\]| Scheme | Marks | AO |
|---|---|---|
| Subtracts their (c) from their (b) Accept (c) − (b) or Writes a sum of odd squares in terms of \(\sum r^2\) and \(\sum r\) PI | M1 | 3.1a |
| Obtains correct expression in terms of \(n\) in any form | A1 | 1.1b |
| Completes a reasoned argument to obtain the correct expression. | R1 | 2.1 |
| (3) | ||
| (10 marks) |