| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| 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 explicit guidance at each step. Part (a) is routine algebraic verification, part (b) applies the telescoping sum technique with clear direction, and part (c) is straightforward substitution. While it requires understanding of the method of differences (a Further Maths topic), the question provides the identity and explicitly tells students which technique to use, making it easier than average overall. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2r-1} - \frac{1}{2r+1} = \frac{2r+1-(2r-1)}{(2r-1)(2r+1)} = \frac{2}{(2r-1)(2r+1)}\) | B1 | Completes a rigorous argument; must include LHS, at least one intermediate step, and RHS |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sum_{r=1}^{n}\frac{2}{(2r-1)(2r+1)} = \sum_{r=1}^{n}\left(\frac{1}{2r-1} - \frac{1}{2r+1}\right)\) | M1 | Writes at least two pairs of subtracting fractions; condone any consistent multiple |
| At least one pair of cancelling fractions shown | M1 | Writes at least one pair of cancelling fractions; condone consistent multiple |
| \(= \frac{1}{1} - \frac{1}{2n+1}= \frac{2n}{2n+1}\) | A1 | Correctly reduces to two terms; PI; condone consistent multiple |
| So \(\sum_{r=1}^{n}\frac{1}{(2r-1)(2r+1)} = \frac{n}{2n+1}\) | R1 | Completes reasoned argument using method of differences; first two pairs and last pair must be shown; accept unsimplified e.g. \(\frac{2n}{4n+2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+\ldots+\frac{1}{99\times101} = \sum_{r=1}^{50}\frac{1}{(2r-1)(2r+1)}\) | M1 | Uses \(n=50\); condone 49 or 51 |
| \(= \frac{50}{101}\) | A1F | Obtains correct exact value; FT their \(\frac{50a}{50b+c}\) |
## Question 7:
### Part 7(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2r-1} - \frac{1}{2r+1} = \frac{2r+1-(2r-1)}{(2r-1)(2r+1)} = \frac{2}{(2r-1)(2r+1)}$ | B1 | Completes a rigorous argument; must include LHS, at least one intermediate step, and RHS |
### Part 7(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum_{r=1}^{n}\frac{2}{(2r-1)(2r+1)} = \sum_{r=1}^{n}\left(\frac{1}{2r-1} - \frac{1}{2r+1}\right)$ | M1 | Writes at least two pairs of subtracting fractions; condone any consistent multiple |
| At least one pair of cancelling fractions shown | M1 | Writes at least one pair of cancelling fractions; condone consistent multiple |
| $= \frac{1}{1} - \frac{1}{2n+1}= \frac{2n}{2n+1}$ | A1 | Correctly reduces to two terms; PI; condone consistent multiple |
| So $\sum_{r=1}^{n}\frac{1}{(2r-1)(2r+1)} = \frac{n}{2n+1}$ | R1 | Completes reasoned argument using method of differences; first two pairs and last pair must be shown; accept unsimplified e.g. $\frac{2n}{4n+2}$ |
### Part 7(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+\ldots+\frac{1}{99\times101} = \sum_{r=1}^{50}\frac{1}{(2r-1)(2r+1)}$ | M1 | Uses $n=50$; condone 49 or 51 |
| $= \frac{50}{101}$ | A1F | Obtains correct exact value; FT their $\frac{50a}{50b+c}$ |7
\begin{enumerate}[label=(\alph*)]
\item Show that, for all integers $r$,
$$\frac { 1 } { 2 r - 1 } - \frac { 1 } { 2 r + 1 } = \frac { 2 } { ( 2 r - 1 ) ( 2 r + 1 ) }$$
7
\item Hence, using the method of differences, show that
$$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 2 r - 1 ) ( 2 r + 1 ) } = \frac { a n } { b n + c }$$
where $a$, $b$ and $c$ are integers to be determined.\\
7
\item Hence, or otherwise, evaluate
$$\frac { 1 } { 1 \times 3 } + \frac { 1 } { 3 \times 5 } + \frac { 1 } { 5 \times 7 } + \ldots + \frac { 1 } { 99 \times 101 }$$
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 2023 Q7 [7]}}