A2 June 2019 Paper 2 Q14
14 Let
\[S_n = \sum_{r=1}^{n} \frac{1}{(r + 1)(r + 3)}\]where \(n \geqslant 1\)
(a) Use the method of differences to show that\[S_n = \frac{5n^2 + an}{12(n + b)(n + c)}\]
where \(a\), \(b\) and \(c\) are integers. [6 marks]
(b) Show that, for any number \(k\) greater than \(\dfrac{12}{5}\), if the difference between \(\dfrac{5}{12}\) and \(S_n\) is less than \(\dfrac{1}{k}\), then\[n \gt \frac{k - 5 + \sqrt{k^2 + 1}}{2}\] [6 marks]
| Scheme | Marks | AO |
|---|---|---|
| Uses partial fractions | M1 | 3.1a |
| Correctly expresses the rational function as partial fractions | A1 | 1.1b |
| Uses the method of differences, showing at least the first three and last two terms (or vice versa) (“Term” here means one fraction minus another fraction) | M1 | 2.5 |
| Correctly uses the method of differences to reduce the expression to four terms (oe) | A1 | 1.1b |
| Correctly expresses their three- or four-term answer with a common denominator | M1 | 1.1a |
| Completes fully correct working to reach the required result | R1 | 2.1 |
Typical solution
\[\frac{1}{r + 1} - \frac{1}{r + 3} = \frac{2}{(r + 1)(r + 3)}\]\[\therefore\ 2S_n = \sum_{r=1}^{n} \frac{1}{r + 1} - \frac{1}{r + 3}\]\[\begin{aligned} &= \frac{1}{2} - \cancel{\frac{1}{4}} \\ &\quad + \frac{1}{3} - \cancel{\frac{1}{5}} \\ &\quad + \cancel{\frac{1}{4}} - \cancel{\frac{1}{6}} \\ &\quad + \cdots \\ &\quad + \cancel{\frac{1}{n - 1}} - \cancel{\frac{1}{n + 1}} \\ &\quad + \cancel{\frac{1}{n}} - \frac{1}{n + 2} \\ &\quad + \cancel{\frac{1}{n + 1}} - \frac{1}{n + 3} \end{aligned}\]\[2S_n = \frac{1}{2} + \frac{1}{3} - \frac{1}{n + 2} - \frac{1}{n + 3}\]\[= \frac{5(n + 2)(n + 3) - 6(n + 3 + n + 2)}{6(n + 2)(n + 3)}\]\[S_n = \frac{5n^2 + 13n}{12(n + 2)(n + 3)}\]| Scheme | Marks | AO |
|---|---|---|
| Writes down a correct inequality. Condone \(\frac{5}{12} - S_n \lt \frac{1}{k}\) | B1 | 3.1a |
| Simplifies the left-hand side of their inequality correctly | M1 | 1.1a |
| Rearranges their inequality to remove fraction, explaining that denominators are positive | M1 | 2.4 |
| Writes their inequality or related equation in simplified quadratic form | M1 | 1.1a |
| Obtains a correct root or roots of the quadratic equation in unsimplified form | M1 | 1.1a |
| Completes a rigorous argument to show the required result. This must include a discussion of the signs of the roots, or other convincing reason that the inequality holds. | R1 | 2.1 |
| (12 marks) |
Typical solution
\[\frac{5}{12} - \frac{5n^2 + 13n}{12(n + 2)(n + 3)} \lt \frac{1}{k}\]\[\frac{5(n + 2)(n + 3) - (5n^2 + 13n)}{12(n + 2)(n + 3)} \lt \frac{1}{k}\]\[\frac{12n + 30}{12(n + 2)(n + 3)} \lt \frac{1}{k}\]\[k(12n + 30) \lt 12(n + 2)(n + 3)\]since both denominators are positive.
\[12n^2 + (60 - 12k)n + (72 - 30k) \gt 0\]\[2n^2 + (10 - 2k)n + (12 - 5k) \gt 0\]The equation
\[2n^2 + (10 - 2k)n + (12 - 5k) = 0\]has a positive and a negative root (since \(12 - 5k \lt 0\))
\[\text{Positive root} = \frac{2k - 10 + \sqrt{(10 - 2k)^2 - 8(12 - 5k)}}{4}\]\[= \frac{2k - 10 + \sqrt{4k^2 + 4}}{4}\]\[= \frac{k - 5 + \sqrt{k^2 + 1}}{2}\]Since the root is positive, if
\[12n^2 + (60 - 12k)n + (72 - 30k) \gt 0\]then
\[n \gt \frac{k - 5 + \sqrt{k^2 + 1}}{2}\]