A2 June 2022 Paper 2 Q4
4 In this question you must show detailed reasoning.
Determine the smallest value of \(n\) for which \(\dfrac{1^2 + 2^2 + \ldots + n^2}{1 + 2 + \ldots + n} \gt 341\). [4]
| Scheme | Marks | AO |
|---|---|---|
| DR \(\dfrac{1^2 + 2^2 + \ldots + n^2}{1 + 2 + \ldots + n} = \dfrac{\frac{1}{6}n(n + 1)(2n + 1)}{\frac{1}{2}n(n + 1)}\) | M1 | 3.1a |
| \(= \dfrac{2n + 1}{3} \Rightarrow \dfrac{2n + 1}{3} \gt 341\) oe | M1 | 2.2a |
| \(\therefore n \gt 511\) | A1 | 1.1 |
| \(\therefore n_{\min} = 512\) | A1 | 3.2a |
| [4] |
Notes
M1: Identifying the two series and quoting the standard series results
M1: Correctly cancelling to set up inequality with two layer fraction(s)
(NB M0 for \(\geqslant 342\)).
If cubic inequality formed then it must be factorised before M1…
A1: …and must be fully solved for A1.
A1: 512 from equation rather than inequality then M1M1A0A1