| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| Session | November |
| 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 Part (a) is a routine surd rationalization (difference of two squares). Part (b) requires recognizing that the given expression telescopes after substituting the result from (a), which is a standard Further Maths technique. Part (c) involves straightforward limit analysis. While this tests multiple concepts and requires careful algebraic manipulation, it follows a well-established pattern for method of differences questions with no novel insight required. |
| Spec | 1.02b Surds: manipulation and rationalising denominators4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| \((x^n - \sqrt{x^{2n}+1})(x^n + \sqrt{x^{2n}+1}) = x^{2n} - (x^{2n}+1) = -1\) | B1 | Uses the difference of two squares and obtains the given answer. AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(u_n = x^{n+1} + \sqrt{x^{2n+2}+1} - x^n - \sqrt{x^{2n}+1}\) | B1 | Uses the identity given in part (a). |
| \(\sum_{n=1}^{N} u_n = x^2 + \sqrt{x^4+1} - x - \sqrt{x^2+1} + x^3 + \sqrt{x^6+1} - x^2 - \sqrt{x^4+1} + \ldots + x^{N+1} + \sqrt{x^{2N+2}+1} - x^N - \sqrt{x^{2N}+1}\) | M1 | Writes at least three complete correct terms, including first and last. |
| \(= x^{N+1} + \sqrt{x^{2N+2}+1} - x - \sqrt{x^2+1}\) | A1 | (Don't allow in terms of \(n\).) |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(-1 < x < 1\) | B1 | |
| \(x^{N+1} \to 0\) as \(n \to \infty \Rightarrow u_1 + u_2 + u_3 + \ldots = 1 - x - \sqrt{x^2+1}\) | M1 A1 | Finds sum to infinity. |
| Total: 3 |
## Question 3:
### Part (a):
| $(x^n - \sqrt{x^{2n}+1})(x^n + \sqrt{x^{2n}+1}) = x^{2n} - (x^{2n}+1) = -1$ | B1 | Uses the difference of two squares and obtains the given answer. AG |
### Part (b):
| $u_n = x^{n+1} + \sqrt{x^{2n+2}+1} - x^n - \sqrt{x^{2n}+1}$ | B1 | Uses the identity given in part (a). |
| $\sum_{n=1}^{N} u_n = x^2 + \sqrt{x^4+1} - x - \sqrt{x^2+1} + x^3 + \sqrt{x^6+1} - x^2 - \sqrt{x^4+1} + \ldots + x^{N+1} + \sqrt{x^{2N+2}+1} - x^N - \sqrt{x^{2N}+1}$ | M1 | Writes at least three complete correct terms, including first and last. |
| $= x^{N+1} + \sqrt{x^{2N+2}+1} - x - \sqrt{x^2+1}$ | A1 | (Don't allow in terms of $n$.) |
| **Total: 3** | | |
### Part (c):
| $-1 < x < 1$ | B1 | |
| $x^{N+1} \to 0$ as $n \to \infty \Rightarrow u_1 + u_2 + u_3 + \ldots = 1 - x - \sqrt{x^2+1}$ | M1 A1 | Finds sum to infinity. |
| **Total: 3** | | |
---3
\begin{enumerate}[label=(\alph*)]
\item By simplifying $\left( x ^ { n } - \sqrt { x ^ { 2 n } + 1 } \right) \left( x ^ { n } + \sqrt { x ^ { 2 n } + 1 } \right)$, show that $\frac { 1 } { x ^ { n } - \sqrt { x ^ { 2 n } + 1 } } = - x ^ { n } - \sqrt { x ^ { 2 n } + 1 }$. [1]\\
Let $u _ { n } = x ^ { n + 1 } + \sqrt { x ^ { 2 n + 2 } + 1 } + \frac { 1 } { x ^ { n } - \sqrt { x ^ { 2 n } + 1 } }$.
\item Use the method of differences to find $\sum _ { \mathrm { n } = 1 } ^ { \mathrm { N } } \mathrm { u } _ { \mathrm { n } }$ in terms of $N$ and $x$.
\item Deduce the set of values of $x$ for which the infinite series
$$u _ { 1 } + u _ { 2 } + u _ { 3 } + \ldots$$
is convergent and give the sum to infinity when this exists.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2020 Q3 [7]}}