| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Infinite series convergence and sum |
| Difficulty | Standard +0.3 This is a standard method of differences question with routine algebraic manipulation. Part (a) is straightforward verification, part (b) applies a well-practiced technique (telescoping series), and part (c) is simple substitution. The question follows a heavily scaffolded structure typical of Further Maths teaching material, requiring no novel insight—just execution of a familiar method. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{r+1}{r+2} - \frac{r}{r+1} = \frac{(r+1)^2 - r(r+2)}{(r+1)(r+2)} = \frac{r^2+2r+1-r^2-2r}{(r+1)(r+2)} = \frac{1}{(r+1)(r+2)}\) | R1 | Must include LHS, at least one intermediate step, and the RHS |
| Answer | Marks | Guidance |
|---|---|---|
| Writes first two pairs (or last two pairs) of corresponding terms of \(\frac{r+1}{r+2}\) and \(\frac{r}{r+1}\) | M1 (1.1a) | — |
| \(\sum_{r=1}^{n} \frac{1}{(r+1)(r+2)} = \sum_{r=1}^{n}\left(\frac{r+1}{r+2} - \frac{r}{r+1}\right)\), writes at least the \(1\)st pair and the \(n\)th pair of corresponding terms and shows pattern of cancelling | M1 (1.1a) | Pattern of cancelling possibly implied by a correct expression |
| Completes to obtain \(\frac{n+1}{n+2} - \frac{1}{2} = \frac{2(n+1)-1(n+2)}{2(n+2)} = \frac{n}{2n+4}\) | R1 (2.1) | Must include \(1\)st and \(n\)th terms and at least one pair of cancelling terms |
| Answer | Marks | Guidance |
|---|---|---|
| Substitutes \(n=2000\) or \(n=1000\) into \(\frac{n}{an+b}\): \(\sum_{r=1}^{2000}\frac{1}{(r+1)(r+2)} = \frac{2000}{2\times2000+4} = \frac{500}{1001}\) | M1 (1.1a) | — |
| Substitutes \(n=2000\) and \(n=1000\) into \(\frac{n}{an+b}\) and subtracts: \(\sum_{r=1}^{1000}\frac{1}{(r+1)(r+2)} = \frac{1000}{2\times1000+4} = \frac{250}{501}\) | M1 (3.1a) | — |
| \(\sum_{r=1001}^{2000}\frac{1}{(r+1)(r+2)} = \frac{500}{1001} - \frac{250}{501} = \frac{250}{501501}\) | A1 (1.1b) | Ignore an approximated answer |
## Question 9(a):
$\frac{r+1}{r+2} - \frac{r}{r+1} = \frac{(r+1)^2 - r(r+2)}{(r+1)(r+2)} = \frac{r^2+2r+1-r^2-2r}{(r+1)(r+2)} = \frac{1}{(r+1)(r+2)}$ | R1 | Must include LHS, at least one intermediate step, and the RHS
---
## Question 9(b):
Writes first two pairs (or last two pairs) of corresponding terms of $\frac{r+1}{r+2}$ and $\frac{r}{r+1}$ | M1 (1.1a) | —
$\sum_{r=1}^{n} \frac{1}{(r+1)(r+2)} = \sum_{r=1}^{n}\left(\frac{r+1}{r+2} - \frac{r}{r+1}\right)$, writes at least the $1$st pair **and** the $n$th pair of corresponding terms and shows pattern of cancelling | M1 (1.1a) | Pattern of cancelling possibly implied by a correct expression
Completes to obtain $\frac{n+1}{n+2} - \frac{1}{2} = \frac{2(n+1)-1(n+2)}{2(n+2)} = \frac{n}{2n+4}$ | R1 (2.1) | Must include $1$st and $n$th terms and at least one pair of cancelling terms
---
## Question 9(c):
Substitutes $n=2000$ **or** $n=1000$ into $\frac{n}{an+b}$: $\sum_{r=1}^{2000}\frac{1}{(r+1)(r+2)} = \frac{2000}{2\times2000+4} = \frac{500}{1001}$ | M1 (1.1a) | —
Substitutes $n=2000$ **and** $n=1000$ into $\frac{n}{an+b}$ and subtracts: $\sum_{r=1}^{1000}\frac{1}{(r+1)(r+2)} = \frac{1000}{2\times1000+4} = \frac{250}{501}$ | M1 (3.1a) | —
$\sum_{r=1001}^{2000}\frac{1}{(r+1)(r+2)} = \frac{500}{1001} - \frac{250}{501} = \frac{250}{501501}$ | A1 (1.1b) | Ignore an approximated answer
---9
\begin{enumerate}[label=(\alph*)]
\item Show that, for all positive integers $r$,
$$\frac { r + 1 } { r + 2 } - \frac { r } { r + 1 } = \frac { 1 } { ( r + 1 ) ( r + 2 ) }$$
9
\item Hence, using the method of differences, show that
$$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( r + 1 ) ( r + 2 ) } = \frac { n } { a n + b }$$
where $a$ and $b$ are integers to be determined.\\
9
\item Hence find the exact value of
$$\sum _ { r = 1001 } ^ { 2000 } \frac { 1 } { ( r + 1 ) ( r + 2 ) }$$
$\_\_\_\_$ The curve $C$ has equation
$$y = \frac { 2 x - 10 } { 3 x - 5 }$$
Figure 1 shows the curve $C$ with its asymptotes.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{47b12ae4-ca3f-472c-9d15-2ef17a2a4d87-12_979_1079_641_468}
\end{center}
\end{figure}
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 2024 Q9 [7]}}