| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Periodic or repeating sequence |
| Difficulty | Moderate -0.3 This is a slightly below-average A-level question. Part (a) requires straightforward substitution into a recurrence relation with basic evaluation of cos(nĪ/2) and (-1)^n. Part (b) exploits periodicity to find a sum, which is a standard technique once the pattern is recognized. The question is computational rather than conceptually demanding, though the periodic structure adds mild interest beyond pure arithmetic sequences. |
| Spec | 1.04e Sequences: nth term and recurrence relations1.04f Sequence types: increasing, decreasing, periodic1.04g Sigma notation: for sums of series1.05a Sine, cosine, tangent: definitions for all arguments |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((u_2=)35+7\cos\!\left(\frac{\pi}{2}\right)-5(-1)^1=40\) | B1* | Correct application with \(n=1\), proceeds correctly to achieve 40 with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u_3=40+7\cos\!\left(\frac{2\pi}{2}\right)-5(-1)^2\ (=28)\) or \(u_4="28"+7\cos\!\left(\frac{3\pi}{2}\right)-5(-1)^3\ (=33)\) | M1 | Correct attempt to use formula to find \(u_3\) or \(u_4\) |
| \(u_3=28\) and \(u_4=33\) | A1 | Both correct values only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((u_5=)35\) | B1 | â |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sum_{r=1}^{25}u_r=6(35+40+"28"+"33")+35\) | M1 | Attempts a correct method; various ways e.g. adding 35 to \(6\times\) sum of four values; attempts to use AP/GP formula score M0 |
| \(=851\) | A1 | Correct answer with no working scores both marks |
## Question 2:
### Part (a)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(u_2=)35+7\cos\!\left(\frac{\pi}{2}\right)-5(-1)^1=40$ | B1* | Correct application with $n=1$, proceeds correctly to achieve 40 with no errors |
### Part (a)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u_3=40+7\cos\!\left(\frac{2\pi}{2}\right)-5(-1)^2\ (=28)$ **or** $u_4="28"+7\cos\!\left(\frac{3\pi}{2}\right)-5(-1)^3\ (=33)$ | M1 | Correct attempt to use formula to find $u_3$ or $u_4$ |
| $u_3=28$ and $u_4=33$ | A1 | Both correct values only |
### Part (b)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(u_5=)35$ | B1 | â |
### Part (b)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum_{r=1}^{25}u_r=6(35+40+"28"+"33")+35$ | M1 | Attempts a correct method; various ways e.g. adding 35 to $6\times$ sum of four values; attempts to use AP/GP formula score M0 |
| $=851$ | A1 | Correct answer with no working scores both marks |
---\begin{enumerate}
\item A sequence $u _ { 1 } , u _ { 2 } , u _ { 3 } \ldots$ is defined by
\end{enumerate}
$$\begin{aligned}
u _ { 1 } & = 35 \\
u _ { n + 1 } & = u _ { n } + 7 \cos \left( \frac { n \pi } { 2 } \right) - 5 ( - 1 ) ^ { n }
\end{aligned}$$
(a) (i) Show that $u _ { 2 } = 40$\\
(ii) Find the value of $u _ { 3 }$ and the value of $u _ { 4 }$
Given that the sequence is periodic with order 4\\
(b) (i) write down the value of $u _ { 5 }$\\
(ii) find the value of $\sum _ { r = 1 } ^ { 25 } u _ { r }$
\hfill \mbox{\textit{Edexcel Paper 2 2023 Q2 [6]}}