| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Integral bounds for series |
| Difficulty | Challenging +1.2 This is a standard Further Maths question on using integral bounds to establish series convergence. It requires recognizing that rectangle areas give the series sum (excluding first term), integrating 1/x³, and combining results—straightforward techniques for FP2 students with no novel insight needed, but more sophisticated than typical A-level Core questions. |
| Spec | 1.08g Integration as limit of sum: Riemann sums4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| or \(\sum_{r=2}^{n}\left(\frac{1}{r}\right)^3\) or \(\sum_{r=1}^{n}\left(\frac{1}{r}\right)^3 - 1\) | M1, A1 | Heights, with at most one extra and/or one omitted; isw. No limits M0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow \sum_{r=1}^{\infty}\left(\frac{1}{r^3}\right) < \frac{1}{2} + 1 = \frac{3}{2}\) | A1, M1, M1, A1 | Integrate correct function: soi. Seen by \(x^2\) in denominator. Or with upper limit of \(n\). www; Compare their answer to (i) (taken to \(\infty\)) with their integral dep on 1st M; Dealing with 1. Dep on previous 2 Ms |
**(i)** Heights of rectangles $= \left(\frac{1}{2}\right)^3, \left(\frac{1}{3}\right)^3, \left(\frac{1}{4}\right)^3, \ldots, \left(\frac{1}{n}\right)^3$
Width of rectangles $= 1$
$\Rightarrow$ Sum of areas $= \left(\frac{1}{2}\right)^3 + \left(\frac{1}{3}\right)^3 + \left(\frac{1}{4}\right)^3 + \ldots + \left(\frac{1}{n}\right)^3$
or $\sum_{r=2}^{n}\left(\frac{1}{r}\right)^3$ or $\sum_{r=1}^{n}\left(\frac{1}{r}\right)^3 - 1$ | M1, A1 | Heights, with at most one extra and/or one omitted; isw. No limits M0
**(ii)** Area $= \int_1^{\infty} \frac{1}{x^3} dx = \left[-\frac{1}{2x^2}\right]_1^{\infty} = \frac{1}{2}$
Since sum of areas of rectangles approximates, but is less than, the area under the curve
$\left(\frac{1}{2}\right)^3 + \left(\frac{1}{3}\right)^3 + \left(\frac{1}{4}\right)^3 + \ldots = \sum_{r=2}^{\infty}\left(\frac{1}{r^3}\right) < \frac{1}{2}$
$\Rightarrow \sum_{r=1}^{\infty}\left(\frac{1}{r^3}\right) < \frac{1}{2} + 1 = \frac{3}{2}$ | A1, M1, M1, A1 | Integrate correct function: soi. Seen by $x^2$ in denominator. Or with upper limit of $n$. www; Compare their answer to (i) (taken to $\infty$) with their integral dep on 1st M; Dealing with 1. Dep on previous 2 Ms
---3 The diagram shows the curve $y = \frac { 1 } { x ^ { 3 } }$ for $1 \leqslant x \leqslant n$ where $n$ is an integer. A set of ( $n - 1$ ) rectangles of unit width is drawn under the curve.\\
\includegraphics[max width=\textwidth, alt={}, center]{736932f1-4007-4a04-a08b-2551db0b136c-2_611_947_1103_557}\\
(i) Write down the sum of the areas of the rectangles.\\
(ii) Hence show that $\sum _ { r = 1 } ^ { \infty } \frac { 1 } { r ^ { 3 } } < \frac { 3 } { 2 }$.
\hfill \mbox{\textit{OCR FP2 2014 Q3 [7]}}