| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Factorial or product method of differences |
| Difficulty | Standard +0.8 This is a Further Maths question requiring students to verify a factorial identity then apply method of differences to a telescoping series. Part (a) is straightforward algebraic manipulation, but part (b) requires recognizing the telescoping pattern and carefully handling the summation limits with factorials, which is more sophisticated than standard A-level series work. |
| Spec | 4.01a Mathematical induction: construct proofs4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{(r-1)!} - \frac{1}{r!} = \frac{r}{r!} - \frac{1}{r!} = \frac{r-1}{r!}\) | B1 | Obtains correct result including at least one intermediate step; all lines must be correct; condone original expression omitted |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sum_{r=1}^{n} \frac{r-1}{r!} = \sum_{r=1}^{n}\left(\frac{1}{(r-1)!} - \frac{1}{r!}\right)\) | M1 | Writes the first two pairs (or last two pairs) of corresponding terms of \(\frac{1}{(r-1)!}\) and \(\frac{1}{r!}\) |
| \(= \frac{1}{0!} - \frac{1}{1!} + \frac{1}{1!} - \frac{1}{2!} + \cdots\) | M1 | Writes at least the first pair and last pair of corresponding terms and shows the pattern of cancelling |
| \(+ \frac{1}{(n-2)!} - \frac{1}{(n-1)!} + \frac{1}{(n-1)!} - \frac{1}{n!}\) | ||
| \(= \frac{1}{0!} - \frac{1}{n!} = 1 - \frac{1}{n!}\) | R1 | Completes fully correct proof with \(a=1\) and \(b=-1\); must include \(1^{\text{st}}\) and \(n^{\text{th}}\) terms and at least one pair of cancelling terms |
## Question 11(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{(r-1)!} - \frac{1}{r!} = \frac{r}{r!} - \frac{1}{r!} = \frac{r-1}{r!}$ | B1 | Obtains correct result including at least one intermediate step; all lines must be correct; condone original expression omitted |
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## Question 11(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum_{r=1}^{n} \frac{r-1}{r!} = \sum_{r=1}^{n}\left(\frac{1}{(r-1)!} - \frac{1}{r!}\right)$ | M1 | Writes the first two pairs (or last two pairs) of corresponding terms of $\frac{1}{(r-1)!}$ and $\frac{1}{r!}$ |
| $= \frac{1}{0!} - \frac{1}{1!} + \frac{1}{1!} - \frac{1}{2!} + \cdots$ | M1 | Writes at least the first pair and last pair of corresponding terms and shows the pattern of cancelling |
| $+ \frac{1}{(n-2)!} - \frac{1}{(n-1)!} + \frac{1}{(n-1)!} - \frac{1}{n!}$ | | |
| $= \frac{1}{0!} - \frac{1}{n!} = 1 - \frac{1}{n!}$ | R1 | Completes fully correct proof with $a=1$ and $b=-1$; must include $1^{\text{st}}$ and $n^{\text{th}}$ terms and at least one pair of cancelling terms |
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\begin{enumerate}[label=(\alph*)]
\item Show that, for all positive integers $r$,
$$\frac { 1 } { ( r - 1 ) ! } - \frac { 1 } { r ! } = \frac { r - 1 } { r ! }$$
⟶\\
11
\item Hence, using the method of differences, show that
$$\sum _ { r = 1 } ^ { n } \frac { r - 1 } { r ! } = a + \frac { b } { n ! }$$
where $a$ and $b$ are integers to be determined.
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 2021 Q11 [4]}}