A2 October 2021 Paper 2 Q4
4. In this question you may assume the results for
\[\sum_{r=1}^{n} r^3, \quad \sum_{r=1}^{n} r^2 \quad \text{and} \quad \sum_{r=1}^{n} r\]The sum of the cubes of 10 consecutive positive odd numbers is 99 800
| Scheme | Marks | AO |
|---|---|---|
| A complete attempt to find the sum of the cubes of the first \(n\) odd numbers using three of the standard summation formulae. Attempts to find \(\sum(2r + 1)^3\) or \(\sum(2r - 1)^3\) by expanding and using summation formulae | M1 | 3.1a |
\(\displaystyle\sum_{r=1}^{n}(2r - 1)^3 = \displaystyle\sum_{r=1}^{n}\left(8r^3 - 12r^2 + 6r - 1\right)\) \(= 8\displaystyle\sum_{r=1}^{n}r^3 - 12\displaystyle\sum_{r=1}^{n}r^2 + 6\displaystyle\sum_{r=1}^{n}r - \displaystyle\sum_{r=1}^{n}1\) or \(\displaystyle\sum_{r=0}^{n-1}(2r + 1)^3 = \displaystyle\sum_{r=0}^{n-1}\left(8r^3 + 12r^2 + 6r + 1\right)\) \(= 8\displaystyle\sum_{r=0}^{n-1}r^3 + 12\displaystyle\sum_{r=0}^{n-1}r^2 + 6\displaystyle\sum_{r=0}^{n-1}r + \displaystyle\sum_{r=0}^{n-1}1\) | M1 | 1.1b |
| \(= 8\dfrac{n^2}{4}(n + 1)^2 - 12\dfrac{n}{6}(n + 1)(2n + 1) + 6\dfrac{n}{2}(n + 1) - n\) or \(= 8\dfrac{(n - 1)^2}{4}(n)^2 + 12\dfrac{(n - 1)}{6}(n)(2n - 1) + 6\dfrac{(n - 1)}{2}(n) + n\) | M1 A1 | 1.1b 1.1b |
| Multiplies out to achieve a correct intermediate line for example \(n(n + 1)\left(2n^2 - 2n + 1\right) - n = 2n^4 - 2n^3 + n^2 + 2n^3 - 2n^2 + n - n\) \(2n^4 + 4n^3 + 2n^2 - 4n^3 - 6n^2 - 2n + 3n^2 + 3n - n\) leading to \(= n^2\left(2n^2 - 1\right)\) cso * | A1* | 2.1 |
| (5) |
Notes
(a)
M1: A complete attempt to find the sum of the cubes of \(n\) odd numbers using three of the standard summation formulae.
M1: Expands \(\sum_{r=1}^{n}(2r - 1)^3\) or \(\sum_{r=0}^{n-1}(2r + 1)^3\) and splits into four appropriate sums.
M1: Applies the result for at least three summations \(\sum_{r=0}^{n-1}r^3, \sum_{r=0}^{n-1}r^2, \sum_{r=0}^{n-1}r\) and \(\sum_{r=0}^{n-1}1\) or \(\sum_{r=1}^{n}r^3, \sum_{r=1}^{n}r^2, \sum_{r=1}^{n}r\) and \(\sum_{r=1}^{n}1\) as appropriate to their expansion provided that there is an attempt at cubing some values.
A1: Correct unsimplified expression.
A1*: Multiplies out to achieve a correct intermediate expression which clearly leads to the correct expression. cso
Special case: If uses \(\sum_{r=1}^{n}(2r + 1)^3\) leading to \(= 8\dfrac{n^2}{4}(n + 1)^2 + 12\dfrac{n}{6}(n + 1)(2n + 1) + 6\dfrac{n}{2}(n + 1) + n\) max score is M1 M0 M1 A1 A0
| Scheme | Marks | AO |
|---|---|---|
\(\displaystyle\sum_{r=n}^{n+9}(2r - 1)^3 = \displaystyle\sum_{r=1}^{n+9}(2r - 1)^3 - \displaystyle\sum_{r=1}^{n-1}(2r - 1)^3\) \(= (n + 9)^2\left(2(n + 9)^2 - 1\right) - (n - 1)^2\left(2(n - 1)^2 - 1\right) = 99800\) or \(\displaystyle\sum_{r=n+1}^{n+10}(2r - 1)^3 = \displaystyle\sum_{r=1}^{n+10}(2r - 1)^3 - \displaystyle\sum_{r=1}^{n}(2r - 1)^3\) \(= (n + 10)^2\left(2(n + 10)^2 - 1\right) - (n)^2\left(2n^2 - 1\right) = 99800\) or \(\displaystyle\sum_{r=n-9}^{n}(2r - 1)^3 = \displaystyle\sum_{r=1}^{n}(2r - 1)^3 - \displaystyle\sum_{r=1}^{n-10}(2r - 1)^3\) \(= (n)^2\left(2(n)^2 - 1\right) - (n - 10)^2\left(2(n - 10)^2 - 1\right) = 99800\) | M1 | 3.1a |
| \(80n^3 + 960n^2 + 5820n - 86760 = 0\) or \(80n^3 + 1200n^2 + 7980n - 79900 = 0\) or \(80n^3 - 1200n^2 + 7980n - 119700 = 0\) | A1 | 1.1b |
| Solves cubic equation | dM1 | 1.1b |
| Achieves \(n = 6\) and the smallest number as 11 or Achieves \(n = 5\) and the smallest number as 11 or Achieves \(n = 15\) and the smallest number as 11 | A1 | 2.3 |
| (4) | ||
| (9 marks) |
Notes
(b)
M1: Uses the answer to part (a) to find the sum of the cubes of the first \(N + 10\) odd numbers minus the sum of the first \(N\) odd numbers and sets equal to 99800 or equivalent.
A1: Correct simplified cubic equation.
dM1: Uses their calculator to solve their cubic equation, dependent on previous method mark.
A1: cao