| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Integral bounds for series |
| Difficulty | Challenging +1.8 This is a Further Maths question requiring integral comparison bounds for series, involving integration of a rational function and careful handling of rectangle positioning. While the integration itself is standard (partial fractions leading to logarithms), the conceptual understanding of upper/lower bounds via rectangles and the multi-step reasoning across two parts elevates this above average difficulty. It's a classic Further Pure technique but requires careful execution and geometric insight. |
| Spec | 1.08g Integration as limit of sum: Riemann sums1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sum_{r=1}^{N}\frac{r}{2r^2-1} = 1 + \sum_{r=2}^{N}\frac{r}{2r^2-1}\) | M1 A1 | Compares with sum of areas of rectangles |
| \(< 1 + \int_1^N \frac{x}{2x^2-1}\,dx\) | M1 | Compares with integral |
| \(\int_1^N \frac{x}{2x^2-1}\,dx = \left[\frac{1}{4}\ln(2x^2-1)\right]_1^N\) | M1 A1 | Finds integral |
| \(\sum_{r=1}^{N}\frac{r}{2r^2-1} < 1 + \frac{1}{4}\ln(2N^2-1)\) | M1 A1 | Inserts limits, AG |
| Total: 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sum_{r=1}^{N}\frac{r}{2r^2-1} = \sum_{r=1}^{N-1}\frac{r}{2r^2-1} + \frac{N}{2N^2-1} > \int_1^N \frac{x}{2x^2-1}\,dx + \frac{N}{2N^2-1}\) | M1 A1 | Compares with integral |
| \(\frac{1}{4}\ln(2N^2-1) + \frac{N}{2N^2-1}\) | A1 | CWO |
| Alternative: \(\sum_{r=1}^{N}\frac{r}{2r^2-1} > \int_1^{N+1}\frac{x}{2x^2-1}\,dx\) | M1 A1 | Compares with integral |
| \(\frac{1}{4}\ln\!\left(2(N+1)^2-1\right)\) | A1 | CWO |
| Total: 3 |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum_{r=1}^{N}\frac{r}{2r^2-1} = 1 + \sum_{r=2}^{N}\frac{r}{2r^2-1}$ | M1 A1 | Compares with sum of areas of rectangles |
| $< 1 + \int_1^N \frac{x}{2x^2-1}\,dx$ | M1 | Compares with integral |
| $\int_1^N \frac{x}{2x^2-1}\,dx = \left[\frac{1}{4}\ln(2x^2-1)\right]_1^N$ | M1 A1 | Finds integral |
| $\sum_{r=1}^{N}\frac{r}{2r^2-1} < 1 + \frac{1}{4}\ln(2N^2-1)$ | M1 A1 | Inserts limits, AG |
| **Total: 7** | | |
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## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum_{r=1}^{N}\frac{r}{2r^2-1} = \sum_{r=1}^{N-1}\frac{r}{2r^2-1} + \frac{N}{2N^2-1} > \int_1^N \frac{x}{2x^2-1}\,dx + \frac{N}{2N^2-1}$ | M1 A1 | Compares with integral |
| $\frac{1}{4}\ln(2N^2-1) + \frac{N}{2N^2-1}$ | A1 | CWO |
| **Alternative:** $\sum_{r=1}^{N}\frac{r}{2r^2-1} > \int_1^{N+1}\frac{x}{2x^2-1}\,dx$ | M1 A1 | Compares with integral |
| $\frac{1}{4}\ln\!\left(2(N+1)^2-1\right)$ | A1 | CWO |
| **Total: 3** | | |
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{e313d6f0-7615-4be5-b13e-2796fd6335e5-04_540_1511_276_274}
The diagram shows the curve $\mathrm { y } = \frac { \mathrm { x } } { 2 \mathrm { x } ^ { 2 } - 1 }$ for $x \geqslant 1$, together with a set of $N - 1$ rectangles of unit\\
width. width.
\begin{enumerate}[label=(\alph*)]
\item By considering the sum of the areas of these rectangles, show that
$$\sum _ { r = 1 } ^ { N } \frac { r } { 2 r ^ { 2 } - 1 } < \frac { 1 } { 4 } \ln \left( 2 N ^ { 2 } - 1 \right) + 1$$
\item Use a similar method to find, in terms of $N$, a lower bound for $\sum _ { r = 1 } ^ { N } \frac { r } { 2 r ^ { 2 } - 1 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2021 Q3 [10]}}