| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2020 |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Rectangle bounds for infinite series |
| Difficulty | Challenging +1.2 This is a standard Further Maths question on using rectangles to bound series via integration. Part (a) requires recognizing that rectangles above the curve give an upper bound and computing the integral, while part (b) mirrors this for a lower bound. The technique is well-established in Further Maths syllabi, requiring careful setup but no novel insight beyond the standard rectangle-integral comparison method. |
| Spec | 1.08g Integration as limit of sum: Riemann sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Sum} = \frac{1}{2^2} + \frac{1}{3^2} + \ldots + \frac{1}{n^2}\) | M1A1 | Heights of rectangles correct |
| \(< \int_1^n \frac{1}{x^2}dx\) | M1 | Compare with integral |
| \(= 1 - \frac{1}{n}\) | M1 | Evaluate integral |
| \(\sum_{r=1}^n \frac{1}{r^2} < 2 - \frac{1}{n} = \frac{2n-1}{n}\) | A1 | AG Deal with the 1 and obtain given answer |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| EITHER Solution 1: Sum area of appropriate set of rectangles | (M1) | Appropriate rectangles |
| \(\frac{1}{1^2} + \frac{1}{2^2} + \ldots + \frac{1}{n^2} > \int_1^{n+1} \frac{1}{x^2}dx\) | M1 | Complete method, obtaining inequality involving sum of rectangle areas and appropriate integral |
| So \(\sum_{r=1}^n \frac{1}{r^2} > 1 - \frac{1}{n+1}\) | A1 | |
| OR Solution 2: Sum area of appropriate set of rectangles | (M1) | Appropriate rectangles |
| \(\frac{1}{1^2} + \frac{1}{2^2} + \ldots + \frac{1}{(n-1)^2} > \int_1^n \frac{1}{x^2}dx = 1 - \frac{1}{n}\) | M1 | Complete method, obtaining inequality involving sum of rectangle areas and appropriate integral |
| So \(\sum_{r=1}^n \frac{1}{r^2} > 1 - \frac{1}{n} + \frac{1}{n^2}\left(= \frac{n^2 - n + 1}{n^2}\right)\) | A1 | |
| Total: 3 |
# Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Sum} = \frac{1}{2^2} + \frac{1}{3^2} + \ldots + \frac{1}{n^2}$ | M1A1 | Heights of rectangles correct |
| $< \int_1^n \frac{1}{x^2}dx$ | M1 | Compare with integral |
| $= 1 - \frac{1}{n}$ | M1 | Evaluate integral |
| $\sum_{r=1}^n \frac{1}{r^2} < 2 - \frac{1}{n} = \frac{2n-1}{n}$ | A1 | AG Deal with the 1 and obtain given answer |
| **Total: 5** | | |
# Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **EITHER** Solution 1: Sum area of appropriate set of rectangles | (M1) | Appropriate rectangles |
| $\frac{1}{1^2} + \frac{1}{2^2} + \ldots + \frac{1}{n^2} > \int_1^{n+1} \frac{1}{x^2}dx$ | M1 | Complete method, obtaining inequality involving sum of rectangle areas and appropriate integral |
| So $\sum_{r=1}^n \frac{1}{r^2} > 1 - \frac{1}{n+1}$ | A1 | |
| **OR** Solution 2: Sum area of appropriate set of rectangles | (M1) | Appropriate rectangles |
| $\frac{1}{1^2} + \frac{1}{2^2} + \ldots + \frac{1}{(n-1)^2} > \int_1^n \frac{1}{x^2}dx = 1 - \frac{1}{n}$ | M1 | Complete method, obtaining inequality involving sum of rectangle areas and appropriate integral |
| So $\sum_{r=1}^n \frac{1}{r^2} > 1 - \frac{1}{n} + \frac{1}{n^2}\left(= \frac{n^2 - n + 1}{n^2}\right)$ | A1 | |
| **Total: 3** | | |4\\
\includegraphics[max width=\textwidth, alt={}, center]{b5503355-3952-47dc-91f4-80a674349b4a-06_538_949_269_557}
The diagram shows the curve with equation $y = \frac { 1 } { x ^ { 2 } }$ for $x > 0$, together with a set of $( n - 1 )$ rectangles of unit width.\\
(a) By considering the sum of the areas of these rectangles, show that
$$\sum _ { r = 1 } ^ { n } \frac { 1 } { r ^ { 2 } } < \frac { 2 n - 1 } { n } .$$
(b) Use a similar method to find, in terms of $n$, a lower bound for $\sum _ { r = 1 } ^ { n } \frac { 1 } { r ^ { 2 } }$.\\
\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q4 [8]}}