| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2023 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Proving standard summation formulae |
| Difficulty | Standard +0.3 Part (a) is a standard method of differences proof for a well-known formula, requiring careful algebraic manipulation but following a prescribed method. Part (b) is a straightforward application of the result from (a) with basic algebraic rearrangement. While this is Further Maths content, it's a routine textbook exercise with no novel insight required. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^34.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(r^2 + 2r + 1 - r^2 = 2r + 1\) | B1 | Expands |
| \(2\sum_{r=1}^{n} r + n = (n+1)^2 - 1^2\) | M1 A1 | Uses method of differences and sums both sides |
| \(\Rightarrow 2\sum_{r=1}^{n} r = n^2 + n = n(n+1)\) | A1 | AG |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sum_{r=1}^{n}(r+a) = \sum_{r=1}^{n} r + an\) | M1 | Relates with \(\sum r\) |
| \(\frac{1}{2}n(n+1) + an = n\) | M1 | Applies \(\sum_{r=1}^{n} r = \frac{1}{2}n(n+1)\) |
| \(a = \frac{1}{2}(1-n)\) | A1 | |
| Total: 3 |
## Question 1:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $r^2 + 2r + 1 - r^2 = 2r + 1$ | **B1** | Expands |
| $2\sum_{r=1}^{n} r + n = (n+1)^2 - 1^2$ | **M1 A1** | Uses method of differences and sums both sides |
| $\Rightarrow 2\sum_{r=1}^{n} r = n^2 + n = n(n+1)$ | **A1** | AG |
| **Total: 4** | | |
---
### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum_{r=1}^{n}(r+a) = \sum_{r=1}^{n} r + an$ | **M1** | Relates with $\sum r$ |
| $\frac{1}{2}n(n+1) + an = n$ | **M1** | Applies $\sum_{r=1}^{n} r = \frac{1}{2}n(n+1)$ |
| $a = \frac{1}{2}(1-n)$ | **A1** | |
| **Total: 3** | | |1
\begin{enumerate}[label=(\alph*)]
\item By considering $( r + 1 ) ^ { 2 } - r ^ { 2 }$, use the method of differences to prove that
$$\sum _ { r = 1 } ^ { n } r = \frac { 1 } { 2 } n ( n + 1 )$$
\item Given that $\sum _ { \mathrm { r } = 1 } ^ { \mathrm { n } } ( \mathrm { r } + \mathrm { a } ) = \mathrm { n }$, find $a$ in terms of $n$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2023 Q1 [7]}}