A2 June 2019 Paper 1 Q4
4. Prove that, for \(n \in \mathbb{Z},\ n \geqslant 0\)
\[\sum_{r=0}^{n} \frac{1}{(r + 1)(r + 2)(r + 3)} = \frac{(n + a)(n + b)}{c(n + 2)(n + 3)}\]where \(a\), \(b\) and \(c\) are integers to be found. (5)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{1}{(r + 1)(r + 2)(r + 3)} \equiv \dfrac{A}{r + 1} + \dfrac{B}{r + 2} + \dfrac{C}{r + 3} \Rightarrow A = \ldots,\ B = \ldots,\ C = \ldots\) \(\left(\text{NB } A = \dfrac{1}{2} \quad B = -1 \quad C = \dfrac{1}{2}\right)\) | M1 | 3.1a |
| \(r = 0\quad \dfrac{1}{2}\left[\dfrac{1}{1} - \dfrac{2}{2} + \dfrac{1}{3}\right]\) or \(\dfrac{1}{2(1)} - \dfrac{1}{2} + \dfrac{1}{2(3)}\) or \(\dfrac{1}{2} - \dfrac{1}{2} + \dfrac{1}{6}\) \(r = 1\quad \dfrac{1}{2}\left[\dfrac{1}{2} - \dfrac{2}{3} + \dfrac{1}{4}\right]\) or \(\dfrac{1}{2(2)} - \dfrac{1}{3} + \dfrac{1}{2(4)}\) or \(\dfrac{1}{4} - \dfrac{1}{3} + \dfrac{1}{8}\) \(r = n - 1\quad \dfrac{1}{2}\left[\dfrac{1}{n} - \dfrac{2}{n + 1} + \dfrac{1}{n + 2}\right]\) or \(\dfrac{1}{2(n)} - \dfrac{1}{n + 1} + \dfrac{1}{2(n + 2)}\) or \(\dfrac{1}{2n} - \dfrac{1}{n + 1} + \dfrac{1}{2n + 4}\) \(r = n\quad \dfrac{1}{2}\left[\dfrac{1}{n + 1} - \dfrac{2}{n + 2} + \dfrac{1}{n + 3}\right]\) or \(\dfrac{1}{2(n + 1)} - \dfrac{1}{n + 2} + \dfrac{1}{2(n + 3)}\) or \(\dfrac{1}{2n + 2} - \dfrac{1}{n + 2} + \dfrac{1}{2n + 6}\) | M1 | 2.1 |
| \(\dfrac{1}{2} - \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{2(n + 2)} - \dfrac{1}{n + 2} + \dfrac{1}{2(n + 3)}\) or \(\dfrac{1}{4} - \dfrac{1}{2(n + 2)} + \dfrac{1}{2(n + 3)}\) | A1 | 1.1b |
| \(= \dfrac{n^2 + 5n + 6 + 2n + 6 - 4n - 12 + 2n + 4}{4(n + 2)(n + 3)}\) | M1 | 1.1b |
| \(= \dfrac{(n + 1)(n + 4)}{4(n + 2)(n + 3)}\) | A1 | 2.2a |
| (5) | ||
| (5 marks) |
Notes
M1: A complete strategy to find \(A\), \(B\) and \(C\) e.g. partial fractions. Allow slip when finding the constant but must be the correct form of partial fractions and correct identity.
M1: Starts the process of differences to identify the relevant fractions at the start and end. Must have attempted a minimum of \(r = 0,\ r = 1,\ \ldots\ r = n - 1\) and \(r = n\)
Follow through on their values of \(A\), \(B\) and \(C\). Look for
\(r = 0 \rightarrow \dfrac{A}{1} - \dfrac{B}{2} + \dfrac{C}{3} \qquad r = 1 \rightarrow \dfrac{A}{2} - \dfrac{B}{3} + \dfrac{C}{4}\)
\(r = n - 1 \rightarrow \dfrac{A}{n} - \dfrac{B}{n + 1} + \dfrac{C}{n + 2} \qquad r = n \rightarrow \dfrac{A}{n + 1} - \dfrac{B}{n + 2} + \dfrac{C}{n + 3}\)
A1: Correct fractions from the beginning and end that do not cancel stated.
M1: Combines all ‘their’ fractions (at least two algebraic fractions) over their correct common denominator, does not need to be the lowest common denominator (allow a slip in the numerator).
A1: Correct answer.
Note: if they start with \(r = 1\) the maximum they can score is M1M0A0M1A0
Note: Proof by induction gains no marks