A2 June 2024 Paper 2 Q13
13
(a) Use the method of differences to show that\[\sum_{r=2}^{n} \frac{1}{(r - 1)r(r + 1)} = \frac{1}{4} - \frac{1}{2n} + \frac{1}{2(n + 1)}\] [5 marks]
(b) Find the smallest integer \(n\) such that\[\sum_{r=2}^{n} \frac{1}{(r - 1)r(r + 1)} \gt 0.24999\] [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Uses partial fractions. | M1 | 3.1a |
| Obtains \(\dfrac{1}{2(r - 1)} - \dfrac{1}{r} + \dfrac{1}{2(r + 1)}\) OE | A1 | 1.1b |
| Writes at least 3 consecutive rows of the sum | M1 | 1.1a |
| Uses the method of differences showing at least the first three and last two terms (or vice versa) | M1 | 2.5 |
| Completes fully correct working to reach the required result. AG | R1 | 2.1 |
| (5) |
Typical solution
\[\frac{1}{(r - 1)r(r + 1)} \equiv \frac{A}{r - 1} + \frac{B}{r} + \frac{C}{r + 1}\]\[1 \equiv Ar(r + 1) + B(r - 1)(r + 1) + C(r - 1)r\]\[r = 0:\quad 1 = -B \Rightarrow B = -1\]\[r = 1:\quad 1 = 2A \Rightarrow A = \tfrac{1}{2}\]\[r = -1:\quad 1 = 2C \Rightarrow C = \tfrac{1}{2}\]\[\begin{aligned} \sum_{r=2}^{n} \frac{1}{(r - 1)r(r + 1)} &= \sum_{r=2}^{n} \left(\frac{1}{2(r - 1)} - \frac{1}{r} + \frac{1}{2(r + 1)}\right) \\ &= \frac{1}{2(1)} - \frac{1}{2} + \cancel{\frac{1}{2(3)}} \\ &\quad + \frac{1}{2(2)} - \cancel{\frac{1}{3}} + \cancel{\frac{1}{2(4)}} \\ &\quad + \cancel{\frac{1}{2(3)}} - \cancel{\frac{1}{4}} + \cancel{\frac{1}{2(5)}} \\ &\quad + \ldots\ldots \\ &\quad + \cancel{\frac{1}{2(n - 3)}} - \cancel{\frac{1}{n - 2}} + \cancel{\frac{1}{2(n - 1)}} \\ &\quad + \cancel{\frac{1}{2(n - 2)}} - \cancel{\frac{1}{n - 1}} + \frac{1}{2n} \\ &\quad + \cancel{\frac{1}{2(n - 1)}} - \frac{1}{n} + \frac{1}{2(n + 1)} \end{aligned}\]\[\begin{aligned} \sum_{r=2}^{n} \frac{1}{(r - 1)r(r + 1)} &= \frac{1}{2} - \frac{1}{2} + \frac{1}{4} + \frac{1}{2n} - \frac{1}{n} + \frac{1}{2(n + 1)} \\ &= \frac{1}{4} - \frac{1}{2n} + \frac{1}{2(n + 1)} \end{aligned}\]| Scheme | Marks | AO |
|---|---|---|
| Obtains an inequality in \(n\) PI | M1 | 3.1a |
| Rearranges their inequality or equation to obtain a quadratic inequality or equation. PI | M1 | 1.1a |
| Deduces the correct value of \(n\) | A1 | 2.2a |
| (3) | ||
| (8 marks) |
Typical solution
\[\sum_{r=2}^{n} \frac{1}{(r - 1)r(r + 1)} \gt 0.24999\]\[\frac{1}{4} - \frac{1}{2n} + \frac{1}{2(n + 1)} \gt 0.24999\]\[0.00001 \gt \frac{1}{2n} - \frac{1}{2(n + 1)}\]\[0.00001 \gt \frac{1}{2n(n + 1)}\]\[2n(n + 1) \gt 10^5\]\[2n^2 + 2n - 10^5 \gt 0\]Solutions to \(2n^2 + 2n - 10^5 = 0\) are 223.1 and \(-224.1\)
\[n \gt 223.1\]\[n = 224\]