| Exam Board | Edexcel |
|---|---|
| Module | PMT Mocks (PMT Mocks) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Periodic or repeating sequence |
| Difficulty | Standard +0.3 Part (a) is a straightforward sigma notation evaluation splitting into geometric series (2^(r-1)), constant sum, and arithmetic series components—all standard A-level techniques. Part (b) requires recognizing the periodic pattern of sin(90n°) which cycles through {1,0,-1,0}, then summing over complete cycles. While (b)(ii) needs careful counting of periods, these are routine manipulations without novel insight. |
| Spec | 1.04f Sequence types: increasing, decreasing, periodic1.04g Sigma notation: for sums of series1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum1.05a Sine, cosine, tangent: definitions for all arguments |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Splits sum: \(\sum_{r=1}^{20} 2^{r-1} - \sum_{r=1}^{20}(3+4r)\) | M1 | Uses a correct method to find the given sum |
| GP sum: \(a=1, r=2\), uses \(S_n = \frac{a(r^n-1)}{r-1}\), giving \(S_{20} = \frac{1(2^{20}-1)}{2-1} = 1048575\) | M1 | Correct method for finding sum of a geometric progression |
| AP sum: \(a=7, d=3\), uses \(S_n = \frac{1}{2}n(a+l)\), giving \(S_{20} = 10(7+83) = 900\) | M1 | Correct method for finding sum of an arithmetic progression |
| \(1048575 - 900 = 1047675\) | A1 | Using correct formula and showing all steps fully leading to \(1047675\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u_1=1,\ u_2=0,\ u_3=-1,\ u_4=0,\ u_5=1,\ldots\) Order 4 | B1 | Obtains terms \(1, 0, -1, 0, 1, 0, \ldots\) and deduces order 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{222}{4} = 55.5\), so 55 complete cycles plus a half cycle; \(55(1+0+(-1)+0) + 1 + 0 = 1\) | M1 | For deducing that the sum can be found using the order of the sequence |
| Answer: \(1\) | A1 | Obtains \(1\) |
## Question 4:
### Part (a): Show that $\sum_{r=1}^{20}(2^{r-1} - 3 - 4r) = 1047675$
| Answer/Working | Mark | Guidance |
|---|---|---|
| Splits sum: $\sum_{r=1}^{20} 2^{r-1} - \sum_{r=1}^{20}(3+4r)$ | M1 | Uses a correct method to find the given sum |
| GP sum: $a=1, r=2$, uses $S_n = \frac{a(r^n-1)}{r-1}$, giving $S_{20} = \frac{1(2^{20}-1)}{2-1} = 1048575$ | M1 | Correct method for finding sum of a geometric progression |
| AP sum: $a=7, d=3$, uses $S_n = \frac{1}{2}n(a+l)$, giving $S_{20} = 10(7+83) = 900$ | M1 | Correct method for finding sum of an arithmetic progression |
| $1048575 - 900 = 1047675$ | A1 | Using correct formula and showing all steps fully leading to $1047675$ |
**(4 marks)**
---
### Part (b)(i): Find the order of the sequence $u_n = \sin(90n°)$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u_1=1,\ u_2=0,\ u_3=-1,\ u_4=0,\ u_5=1,\ldots$ Order **4** | B1 | Obtains terms $1, 0, -1, 0, 1, 0, \ldots$ and deduces order 4 |
**(1 mark)**
---
### Part (b)(ii): Find $\sum_{r=1}^{222} u_r$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{222}{4} = 55.5$, so 55 complete cycles plus a half cycle; $55(1+0+(-1)+0) + 1 + 0 = 1$ | M1 | For deducing that the sum can be found using the order of the sequence |
| Answer: $1$ | A1 | Obtains $1$ |
**(2 marks)**
---4. (a) Show that $\sum _ { r = 1 } ^ { 20 } \left( 2 ^ { r - 1 } - 3 - 4 r \right) = 1047675$\\
(b) A sequence has $n$th term $u _ { n } = \sin \left( 90 n ^ { \circ } \right) n \geq 1$
\begin{enumerate}[label=(\roman*)]
\item Find the order of the sequence.
\item Find $\sum _ { r = 1 } ^ { 222 } u _ { r }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel PMT Mocks Q4 [7]}}