June 2023 Paper 3 Q13
13
The questions in this section refer to the article on the Insert. You should read the article before attempting the questions.
The relevant parts of the article “Approximating series” are reproduced below; the line numbers are those printed on the Insert.
Lines 4–5
The sum of the squares of the first \(n\) natural numbers, \(1^2 + 2^2 + 3^2 + \ldots + n^2\), can be expressed exactly as a formula, \(\displaystyle\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}\).Line 10
Euler’s approximate summation formulaLines 11–13
In 1741, the mathematician Leonhard Euler published an approximate formula for summing a series. In modern notation, this can be expressed as follows.
\(\displaystyle\sum_{r=1}^{n}\mathrm{f}(r) \approx \int_1^n \mathrm{f}(x)\,\mathrm{d}x + \frac{\mathrm{f}(n) + \mathrm{f}(1)}{2} + \frac{\mathrm{f}(1) - \mathrm{f}(2)}{12} - \frac{\mathrm{f}(n) - \mathrm{f}(n+1)}{12}\)
Prove that Euler’s approximate formula, as given in line 13, when applied to \(\displaystyle\sum_{r=1}^{n} r^2\) gives exactly \(\dfrac{n(n+1)(2n+1)}{6}\). [4]
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\int_1^n x^2\,\mathrm{d}x + \frac{n^2 + 1^2}{2} + \frac{1^2 - 2^2}{12} - \frac{n^2 - (n+1)^2}{12}\) | B1 | 1.1a |
| \(\left[\dfrac{x^3}{3}\right]_1^n + \dfrac{6n^2 + 6 + 1 - 4 - n^2 + n^2 + 2n + 1}{12}\) \(\dfrac{n^3 - 1}{3} + \left[\dfrac{6n^2 + 2n + 4}{12}\right]\) | B1 | 1.1 |
| \(\dfrac{2n^3 + 3n^2 + n}{6}\) | B1 | 2.1 |
| \(\dfrac{n\left(2n^2 + 3n + 1\right)}{6} = \dfrac{n(n+1)(2n+1)}{6}\) | B1 | 2.2a |
| [4] |
Notes
B1: Substitute correctly into formula
Condone \(r\) but not \(n\) instead of \(x\)
B1: Integral correctly evaluated
B1: Correct single fraction – may not be fully simplified
OR \(\dfrac{n(n+1)(2n+1)}{6} = \dfrac{2n^3 + 3n^2 + n}{6}\)
B1: Correct completion
Intermediate step needed
Dep on B3
OR convincing comparison of \(\dfrac{n(n+1)(2n+1)}{6}\) and \(\dfrac{n^3 - 1}{3} + \dfrac{6n^2 + 2n + 4}{12}\)