| Exam Board | WJEC |
|---|---|
| Module | Further Unit 1 (Further Unit 1) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Partial fractions then method of differences |
| Difficulty | Standard +0.3 This is a standard telescoping series question with straightforward algebraic manipulation. Part (a) is routine verification (1 mark), part (b) requires recognizing the telescoping pattern and simplifying—a common Further Maths technique but mechanical once spotted, and part (c) is trivial observation about division by zero at r=1. The question is slightly above average difficulty due to being Further Maths content, but it follows a well-established template with no novel insight required. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \(\frac{2}{n-1} - \frac{2}{n+1} = \frac{2(n+1)-2(n-1)}{n^2-1} = \frac{4}{n^2-1}\) | B1 | Convincing |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \(\sum_{r=2}^{n} \frac{4}{r^2-1}\) | ||
| Answer: \(= \left(2 - \frac{2}{3}\right) + \left(1 - \frac{2}{4}\right) + \left(\frac{2}{3} - \frac{2}{5}\right) + \left(\frac{2}{4} - \frac{2}{6}\right) + \ldots + \left(\frac{2}{n-3} - \frac{2}{n-1}\right) + \left(\frac{2}{n-2} - \frac{2}{n}\right) + \left(\frac{2}{n-1} - \frac{2}{n+1}\right)\) | M1 A1 | 3 correct brackets. Correct 1st, last and one other bracket |
| Answer: \(= 2 + 1 - \frac{2}{n} - \frac{2}{n+1}\) | A1 | si |
| Answer: \(= 3 - \frac{2(n+1)+2n}{n(n+1)}\) | ||
| Answer: \(= \frac{3n(n+1)-2(n+1)-2n}{n(n+1)}\) | m1 | For common algebraic denominator |
| Answer: \(= \frac{3n^2-n-2}{n(n+1)}\) | A1 | Correct simplification of their numerator (at least quadratic) |
| Answer: \(= \frac{(3n+2)(n-1)}{n(n+1)}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: The 1st term is undefined | B1 | oe |
**Part a)**
Answer: $\frac{2}{n-1} - \frac{2}{n+1} = \frac{2(n+1)-2(n-1)}{n^2-1} = \frac{4}{n^2-1}$ | B1 | Convincing
**Part b)**
Answer: $\sum_{r=2}^{n} \frac{4}{r^2-1}$ |
Answer: $= \left(2 - \frac{2}{3}\right) + \left(1 - \frac{2}{4}\right) + \left(\frac{2}{3} - \frac{2}{5}\right) + \left(\frac{2}{4} - \frac{2}{6}\right) + \ldots + \left(\frac{2}{n-3} - \frac{2}{n-1}\right) + \left(\frac{2}{n-2} - \frac{2}{n}\right) + \left(\frac{2}{n-1} - \frac{2}{n+1}\right)$ | M1 A1 | 3 correct brackets. Correct 1st, last and one other bracket
Answer: $= 2 + 1 - \frac{2}{n} - \frac{2}{n+1}$ | A1 | si
Answer: $= 3 - \frac{2(n+1)+2n}{n(n+1)}$ |
Answer: $= \frac{3n(n+1)-2(n+1)-2n}{n(n+1)}$ | m1 | For common algebraic denominator
Answer: $= \frac{3n^2-n-2}{n(n+1)}$ | A1 | Correct simplification of their numerator (at least quadratic)
Answer: $= \frac{(3n+2)(n-1)}{n(n+1)}$ | A1 | cao
**Part c)**
Answer: The 1st term is undefined | B1 | oe\begin{enumerate}[label=(\alph*)]
\item Show that $\frac{2}{n-1} - \frac{2}{n+1}$ can be expressed as $\frac{4}{(n^2-1)}$. [1]
\item Hence, find an expression for $\sum_{r=2}^{n} \frac{4}{(r^2-1)}$ in the form $\frac{(an+b)(n+c)}{n(n+1)}$,
where $a$, $b$, $c$ are integers whose values are to be determined. [6]
\item Explain why $\sum_{r=1}^{100} \frac{4}{(r^2-1)}$ cannot be calculated. [1]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 1 2018 Q5 [8]}}