A2 June 2024 Paper 2 Q4
4. Use the method of differences to show that
\[\sum_{r=1}^{n} \frac{2}{(r + 4)(r + 6)} = \frac{n(an + b)}{30(n + 5)(n + 6)}\]where \(a\) and \(b\) are integers to be determined.
(6)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{2}{(r + 4)(r + 6)} \equiv \dfrac{A}{r + 4} + \dfrac{B}{r + 6} \Rightarrow A = \ldots,\ B = \ldots\) | M1 | 3.1a |
| \(\dfrac{2}{(r + 4)(r + 6)} \equiv \dfrac{1}{r + 4} - \dfrac{1}{r + 6}\) | A1 | 1.1b |
| \(\displaystyle\sum_{r=1}^{n} \frac{2}{(r + 4)(r + 6)} = \sum_{r=1}^{n} \frac{1}{r + 4} - \frac{1}{r + 6}\) \(\dfrac{1}{5} - \dfrac{1}{7} + \dfrac{1}{6} - \dfrac{1}{8} + \dfrac{1}{7} - \dfrac{1}{9} + \ldots\) \(+ \dfrac{1}{n + 2} - \dfrac{1}{n + 4} + \dfrac{1}{n + 3} - \dfrac{1}{n + 5} + \dfrac{1}{n + 4} - \dfrac{1}{n + 6}\) | M1 | 2.1 |
| \(= \dfrac{1}{5} + \dfrac{1}{6} - \dfrac{1}{n + 5} - \dfrac{1}{n + 6}\) | A1 | 2.2a |
| \(= \dfrac{1}{5} + \dfrac{1}{6} - \dfrac{1}{n + 5} - \dfrac{1}{n + 6} = \dfrac{11(n + 5)(n + 6) - 30(n + 6) - 30(n + 5)}{30(n + 5)(n + 6)}\) | M1 | 1.1b |
| \(= \dfrac{n(11n + 61)}{30(n + 5)(n + 6)}\) | A1 | 1.1b |
| (6) | ||
| (6 marks) |
Notes
M1: Recognises the need to find partial fractions and applies a correct method leading to finding values for \(A\) and \(B\)
Allow a slip when finding the constants
A1: Correct partial fractions seen at any stage. Not just values for \(A\) and \(B\) listed
Note: Proof by induction will not score the next 4 marks.
M1: Starts the process of finding terms at the start and at the end, in order to establish the non-cancelling terms
Must have attempted a minimum of \(r = 1,\ r = 2,\ \ldots\ r = n - 1\) and \(r = n\), this may be implied by their correct non-cancelling terms.
Follow through on their values of \(A\) and \(B\). Look for
\(r = 1 \rightarrow \dfrac{A}{5} - \dfrac{B}{7}\) \(r = 2 \rightarrow \dfrac{A}{6} - \dfrac{B}{8}\)
\(r = n - 1 \rightarrow \dfrac{A}{n + 3} - \dfrac{B}{n + 5}\) \(r = n \rightarrow \dfrac{A}{n + 4} - \dfrac{B}{n + 6}\)
A1: Correct non-cancelling terms which may be listed separately.
Correct fractions from the beginning and end that do not cancel stated.
M1: Combines ‘their’ fractions of the form \(p + \dfrac{q}{n + 5} + \dfrac{r}{n + 6}\) over a correct common denominator which does not need to be the lowest common denominator and obtains a quadratic expression in the numerator.
A1: Correct answer.
Note: If they start with \(r = 0\) the maximum they can score is M1A1M0A0M1A0