| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2024 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Surd rationalization method of differences |
| Difficulty | Challenging +1.2 This is a structured Further Maths question requiring surd rationalization (a standard technique), telescoping series recognition, and algebraic manipulation. Part (a) is routine rationalization, part (b) is standard method of differences application, and part (c) requires equating expressions and solving for k. While it involves multiple steps and Further Maths content, each component follows well-established techniques without requiring novel insight. |
| Spec | 1.02b Surds: manipulation and rationalising denominators4.06a Summation formulae: sum of r, r^2, r^34.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(\frac{r}{\sqrt{r(r+1)}+\sqrt{r(r-1)}} \times \frac{\sqrt{r(r+1)}-\sqrt{r(r-1)}}{\sqrt{r(r+1)}-\sqrt{r(r-1)}}\) | M1 | A correct multiplier to rationalise the denominator |
| \(= \frac{r\left(\sqrt{r(r+1)}-\sqrt{r(r-1)}\right)}{r(r+1)-r(r-1)} = \frac{\sqrt{r(r+1)}-\sqrt{r(r-1)}}{2}\) or \(A = \frac{1}{2}\) | A1 | Correct expression or correct value for \(A\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| Applies method of differences for \(r=1\) and \(r=n\) | M1 | Obtains one correct row |
| Applies method of differences for \(r=1\), \(r=n\) and either \(r=2\) or \(r=n-1\) | M1 | Obtains 2 correct rows |
| \(= \frac{1}{2}\sqrt{n(n+1)}\) oe e.g. \(\frac{\sqrt{n^2+n}}{2}\) | A1 | Correct expression in terms of \(n\). No incorrect terms. No "0" so \(\frac{1}{2}\left(\sqrt{n(n+1)}-0\right)\) is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(\sum r = \frac{1}{2}n(n+1)\) stated or used | M1 | States or uses correct summation formula for integers |
| \(\frac{k}{2}\sqrt{n(n+1)} = \sqrt{\frac{1}{2}n(n+1)} \Rightarrow \frac{k}{2} = \sqrt{\frac{1}{2}} \Rightarrow k = \sqrt{2}\) | A1 | \(\sqrt{2}\) only (not \(\pm\)). \(k=\sqrt{2}\) must not come from a clearly incorrect equation |
# Question 3:
## Part 3(a):
| Working | Mark | Notes |
|---------|------|-------|
| $\frac{r}{\sqrt{r(r+1)}+\sqrt{r(r-1)}} \times \frac{\sqrt{r(r+1)}-\sqrt{r(r-1)}}{\sqrt{r(r+1)}-\sqrt{r(r-1)}}$ | M1 | A correct multiplier to rationalise the denominator |
| $= \frac{r\left(\sqrt{r(r+1)}-\sqrt{r(r-1)}\right)}{r(r+1)-r(r-1)} = \frac{\sqrt{r(r+1)}-\sqrt{r(r-1)}}{2}$ or $A = \frac{1}{2}$ | A1 | Correct expression or correct value for $A$ |
## Part 3(b):
| Working | Mark | Notes |
|---------|------|-------|
| Applies method of differences for $r=1$ and $r=n$ | M1 | Obtains one correct row |
| Applies method of differences for $r=1$, $r=n$ and either $r=2$ or $r=n-1$ | M1 | Obtains 2 correct rows |
| $= \frac{1}{2}\sqrt{n(n+1)}$ oe e.g. $\frac{\sqrt{n^2+n}}{2}$ | A1 | Correct expression in terms of $n$. No incorrect terms. No "0" so $\frac{1}{2}\left(\sqrt{n(n+1)}-0\right)$ is A0 |
## Part 3(c):
| Working | Mark | Notes |
|---------|------|-------|
| $\sum r = \frac{1}{2}n(n+1)$ stated or used | M1 | States or uses correct summation formula for integers |
| $\frac{k}{2}\sqrt{n(n+1)} = \sqrt{\frac{1}{2}n(n+1)} \Rightarrow \frac{k}{2} = \sqrt{\frac{1}{2}} \Rightarrow k = \sqrt{2}$ | A1 | $\sqrt{2}$ only (not $\pm$). $k=\sqrt{2}$ must not come from a clearly incorrect equation |
---\begin{enumerate}
\item (a) Show that for $r \geqslant 1$
\end{enumerate}
$$\frac { r } { \sqrt { r ( r + 1 ) } + \sqrt { r ( r - 1 ) } } \equiv A ( \sqrt { r ( r + 1 ) } - \sqrt { r ( r - 1 ) } )$$
where $A$ is a constant to be determined.\\
(b) Hence use the method of differences to determine a simplified expression for
$$\sum _ { r = 1 } ^ { n } \frac { r } { \sqrt { r ( r + 1 ) } + \sqrt { r ( r - 1 ) } }$$
(c) Determine, as a surd in simplest form, the constant $k$ such that
$$\sum _ { r = 1 } ^ { n } \frac { k r } { \sqrt { r ( r + 1 ) } + \sqrt { r ( r - 1 ) } } = \sqrt { \sum _ { r = 1 } ^ { n } r }$$
\hfill \mbox{\textit{Edexcel F2 2024 Q3 [7]}}