| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Method of differences with given identity |
| Difficulty | Standard +0.3 This is a standard method of differences question with the identity provided. Part (a) requires algebraic verification of a given result, part (b) applies telescoping summation (a core Further Maths technique), and part (c) evaluates a simple limit. While it's a Further Maths topic, the question is routine and follows a well-established template with no novel insight required. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(r-1) - f(r) = \frac{1}{r(r+1)} - \frac{1}{(r+1)(r+2)}\) | M1 | |
| \(= \frac{r+2-r}{r(r+1)(r+2)} = \frac{2}{r(r+1)(r+2)}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sum_{r=1}^{n} \frac{1}{r(r+1)(r+2)} = \frac{1}{2}\left(\frac{1}{1\times2} - \frac{1}{2\times3}\right) + \ldots + \frac{1}{2}\left(\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)}\right)\) | M1A1 | Sets up method of differences |
| \(= \frac{1}{4} - \frac{1}{2}\left(\frac{1}{(n+1)(n+2)}\right)\) | B1 | OE Shows cancellation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sum_{r=1}^{\infty} \frac{1}{r(r+1)(r+2)} = \frac{1}{4}\) | B1FT | Follow through from their part (b) |
## Question 1(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(r-1) - f(r) = \frac{1}{r(r+1)} - \frac{1}{(r+1)(r+2)}$ | M1 | |
| $= \frac{r+2-r}{r(r+1)(r+2)} = \frac{2}{r(r+1)(r+2)}$ | A1 | AG |
---
## Question 1(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum_{r=1}^{n} \frac{1}{r(r+1)(r+2)} = \frac{1}{2}\left(\frac{1}{1\times2} - \frac{1}{2\times3}\right) + \ldots + \frac{1}{2}\left(\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)}\right)$ | M1A1 | Sets up method of differences |
| $= \frac{1}{4} - \frac{1}{2}\left(\frac{1}{(n+1)(n+2)}\right)$ | B1 | OE Shows cancellation |
---
## Question 1(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum_{r=1}^{\infty} \frac{1}{r(r+1)(r+2)} = \frac{1}{4}$ | B1FT | Follow through from their part (b) |
---1 (a) Given that $\mathrm { f } ( r ) = \frac { 1 } { ( r + 1 ) ( r + 2 ) }$, show that
$$\mathrm { f } ( r - 1 ) - \mathrm { f } ( r ) = \frac { 2 } { r ( r + 1 ) ( r + 2 ) } .$$
(b) Hence find $\sum _ { r = 1 } ^ { n } \frac { 1 } { r ( r + 1 ) ( r + 2 ) }$.\\
(c) Deduce the value of $\sum _ { r = 1 } ^ { \infty } \frac { 1 } { r ( r + 1 ) ( r + 2 ) }$.\\
\hfill \mbox{\textit{CAIE Further Paper 1 2020 Q1 [6]}}