| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Surd rationalization method of differences |
| Difficulty | Standard +0.8 This is a Further Maths question combining surd rationalization with method of differences. Part (a) requires multiplying by conjugate (standard technique), part (b) requires recognizing telescoping series pattern, and part (c) involves careful algebraic manipulation to express the result in the required form. While each technique is standard for Further Maths students, the combination and the need for precise algebraic manipulation across multiple parts elevates this slightly above average difficulty. |
| Spec | 1.02b Surds: manipulation and rationalising denominators4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{2}{\sqrt{r}+\sqrt{r-2}} \times \frac{\sqrt{r}-\sqrt{r-2}}{\sqrt{r}-\sqrt{r-2}}\) | M1 | Intention to multiply by correct fraction; may use \(\frac{\sqrt{r-2}-\sqrt{r}}{\sqrt{r-2}-\sqrt{r}}\); "2" may be missing |
| \(\frac{2(\sqrt{r}-\sqrt{r-2})}{r-(r-2)} = \sqrt{r}-\sqrt{r-2}\) | A1* | Fully correct proof; result from rationalisation that is not the given answer must be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(r=2\): \(\sqrt{2}-\sqrt{0}\); \(r=3\): \(\sqrt{3}-1\); \(r=4\): \(2-\sqrt{2}\); ... \(r=n-2\): \(\sqrt{n-2}-\sqrt{n-4}\); \(r=n-1\): \(\sqrt{n-1}-\sqrt{n-3}\); \(r=n\): \(\sqrt{n}-\sqrt{n-2}\) | M1 | Correct process of differences; at least three of six expressions; evidence of at least one cancelling pair |
| \(\sum_{r=2}^{n} \frac{2}{\sqrt{r}+\sqrt{r-2}} = \sqrt{n}+\sqrt{n-1}-1\) | A1 | Correct algebraic terms or correct constant term(s); accept \(-\sqrt{1}-\sqrt{0}\) |
| A1 | Fully correct simplified expression; must simplify \(\sqrt{1}\) to 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sum_{r=4}^{50} = f(50)-f(3) = \sqrt{50}+\sqrt{49}-1-(\sqrt{3}+\sqrt{2}-1)\) | M1 | Attempts \(f(50)-f(3)\) using answer to (b); must indicate subtraction |
| \(= 5\sqrt{2}+7-1-\sqrt{3}-\sqrt{2}+1 = 7+4\sqrt{2}-\sqrt{3}\) | A1 | Correct expression; \(A=7\), \(B=4\), \(C=-1\) following correct answer to (b) |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{2}{\sqrt{r}+\sqrt{r-2}} \times \frac{\sqrt{r}-\sqrt{r-2}}{\sqrt{r}-\sqrt{r-2}}$ | M1 | Intention to multiply by correct fraction; may use $\frac{\sqrt{r-2}-\sqrt{r}}{\sqrt{r-2}-\sqrt{r}}$; "2" may be missing |
| $\frac{2(\sqrt{r}-\sqrt{r-2})}{r-(r-2)} = \sqrt{r}-\sqrt{r-2}$ | A1* | Fully correct proof; result from rationalisation that is not the given answer must be seen |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $r=2$: $\sqrt{2}-\sqrt{0}$; $r=3$: $\sqrt{3}-1$; $r=4$: $2-\sqrt{2}$; ... $r=n-2$: $\sqrt{n-2}-\sqrt{n-4}$; $r=n-1$: $\sqrt{n-1}-\sqrt{n-3}$; $r=n$: $\sqrt{n}-\sqrt{n-2}$ | M1 | Correct process of differences; at least three of six expressions; evidence of at least one cancelling pair |
| $\sum_{r=2}^{n} \frac{2}{\sqrt{r}+\sqrt{r-2}} = \sqrt{n}+\sqrt{n-1}-1$ | A1 | Correct algebraic terms **or** correct constant term(s); accept $-\sqrt{1}-\sqrt{0}$ |
| | A1 | Fully correct simplified expression; must simplify $\sqrt{1}$ to 1 |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum_{r=4}^{50} = f(50)-f(3) = \sqrt{50}+\sqrt{49}-1-(\sqrt{3}+\sqrt{2}-1)$ | M1 | Attempts $f(50)-f(3)$ using answer to (b); must indicate subtraction |
| $= 5\sqrt{2}+7-1-\sqrt{3}-\sqrt{2}+1 = 7+4\sqrt{2}-\sqrt{3}$ | A1 | Correct expression; $A=7$, $B=4$, $C=-1$ following correct answer to (b) |
---\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
Solutions relying on calculator technology are not acceptable.\\
(a) Show that, for $r \geqslant 2$
$$\frac { 2 } { \sqrt { r } + \sqrt { r - 2 } } = \sqrt { r } - \sqrt { r - 2 }$$
(b) Hence use the method of differences to determine
$$\sum _ { r = 2 } ^ { n } \frac { 2 } { \sqrt { r } + \sqrt { r - 2 } }$$
giving your answer in simplest form.\\
(c) Hence show that
$$\sum _ { r = 4 } ^ { 50 } \frac { 2 } { \sqrt { r } + \sqrt { r - 2 } } = A + B \sqrt { 2 } + C \sqrt { 3 }$$
where $A$, $B$ and $C$ are integers to be determined.
\hfill \mbox{\textit{Edexcel F2 2023 Q1 [7]}}