| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2020 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | Periodic Sequences |
| Difficulty | Challenging +1.8 This is a challenging Further Maths question requiring understanding of periodic sequences, algebraic manipulation with the periodicity condition (a₄ = a₁), and pattern recognition for summation. While the individual steps are methodical, the setup requires insight into how periodicity constrains the recurrence relation, and part (c) demands recognizing the repeating cycle structure. This is above-average difficulty but not exceptionally hard for Further Maths students who understand periodic sequences. |
| Spec | 1.04e Sequences: nth term and recurrence relations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses \(a_{n+1} = \frac{k(a_n + 2)}{a_n}\) once with \(a_1 = 2\) | M1 | |
| \((a_1=2),\ a_2 = 2k,\ a_3 = k+1,\ a_4 = \frac{k(k+3)}{k+1}\). Finds four consecutive terms and sets \(a_4 = a_1\) | M1 | |
| \(\frac{k(k+3)}{k+1} = 2 \Rightarrow k^2 + 3k = 2k+2 \Rightarrow k^2 + k - 2 = 0\) | A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States that when \(k=1\), all terms are the same and concludes the sequence does not have a period of order 3 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Deduces the repeating terms are \(a_{1/4} = 2,\ a_{2/5} = -4,\ a_{3/6} = -1\) | B1 | |
| \(\sum_{n=1}^{80} a_k = 26\times(2 + -4 + -1) + 2 + -4\) | M1 | |
| \(= -80\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Applies sequence formula \(a_{n+1} = \frac{k(a_n + 2)}{a_n}\) | M1 | Usually scored by finding second term, e.g. \(a_2 = 2k\) or \(a_{1+1} = \frac{k(2+2)}{2}\) |
| Attempts to find \(a_1 \to a_4\), sets \(a_1 = a_4\) | M1 | Condone slips. Other methods available, e.g. set \(a_4 = 2\), work backwards to find \(a_3\) and equate to \(k+1\). No requirement to see \(a_1\) or labels. |
| FYI: \(a_1=2,\ a_2=2k,\ a_3=k+1,\ a_4=\frac{k(k+3)}{k+1}\), so \(2=\frac{k(k+3)}{k+1}\) | ||
| Proceeds to given answer with all necessary working shown | A1* | All necessary lines required |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| States that when \(k=1\), all terms are the same and concludes the sequence does not have period of order 3 | B1 | Do not accept "the terms just repeat." Must reference it does not have order 3. Accept "it has order 1." Acceptable to state \(a_2 = a_1 = 2\) and state sequence does not have order 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Deduces repeating terms are \(a_{1/4}=2,\ a_{2/5}=-4,\ a_{3/6}=-1\) | B1 | |
| Uses clear strategy to find sum to 80 terms using multiples of first three terms | M1 | e.g. \(\sum_{r=1}^{80}a_r = \left(\sum_{r=1}^{78}a_r\right)+a_{79}+a_{80} = 26\times(2+-4+-1)+2+-4\) or \(\sum_{r=1}^{80}a_r = \left(\sum_{r=1}^{81}a_r\right)-a_{81} = 27\times(2+-4+-1)-(-1)\) |
| For candidates using \(k\): award for \(27\times2 + 27\times(2k) + 26\times(k+1)\) or \(80k+80\) | If candidates substitute \(k=-2\) into \(80k+80\) to get \(-80\), all 3 marks awarded | |
| \(-80\) | A1 |
## Question 13:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $a_{n+1} = \frac{k(a_n + 2)}{a_n}$ once with $a_1 = 2$ | M1 | |
| $(a_1=2),\ a_2 = 2k,\ a_3 = k+1,\ a_4 = \frac{k(k+3)}{k+1}$. Finds four consecutive terms and sets $a_4 = a_1$ | M1 | |
| $\frac{k(k+3)}{k+1} = 2 \Rightarrow k^2 + 3k = 2k+2 \Rightarrow k^2 + k - 2 = 0$ | A1* | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States that when $k=1$, all terms are the same and concludes the sequence does not have a period of order 3 | B1 | |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Deduces the repeating terms are $a_{1/4} = 2,\ a_{2/5} = -4,\ a_{3/6} = -1$ | B1 | |
| $\sum_{n=1}^{80} a_k = 26\times(2 + -4 + -1) + 2 + -4$ | M1 | |
| $= -80$ | A1 | |
# Question 13 (Sequence):
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Applies sequence formula $a_{n+1} = \frac{k(a_n + 2)}{a_n}$ | M1 | Usually scored by finding second term, e.g. $a_2 = 2k$ or $a_{1+1} = \frac{k(2+2)}{2}$ |
| Attempts to find $a_1 \to a_4$, sets $a_1 = a_4$ | M1 | Condone slips. Other methods available, e.g. set $a_4 = 2$, work backwards to find $a_3$ and equate to $k+1$. No requirement to see $a_1$ or labels. |
| FYI: $a_1=2,\ a_2=2k,\ a_3=k+1,\ a_4=\frac{k(k+3)}{k+1}$, so $2=\frac{k(k+3)}{k+1}$ | | |
| Proceeds to given answer with all necessary working shown | A1* | All necessary lines required |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| States that when $k=1$, all terms are the same and concludes the sequence does not have period of order 3 | B1 | Do not accept "the terms just repeat." Must reference it does not have order 3. Accept "it has order 1." Acceptable to state $a_2 = a_1 = 2$ and state sequence does not have order 3 |
## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Deduces repeating terms are $a_{1/4}=2,\ a_{2/5}=-4,\ a_{3/6}=-1$ | B1 | |
| Uses clear strategy to find sum to 80 terms using multiples of first three terms | M1 | e.g. $\sum_{r=1}^{80}a_r = \left(\sum_{r=1}^{78}a_r\right)+a_{79}+a_{80} = 26\times(2+-4+-1)+2+-4$ or $\sum_{r=1}^{80}a_r = \left(\sum_{r=1}^{81}a_r\right)-a_{81} = 27\times(2+-4+-1)-(-1)$ |
| For candidates using $k$: award for $27\times2 + 27\times(2k) + 26\times(k+1)$ or $80k+80$ | | If candidates substitute $k=-2$ into $80k+80$ to get $-80$, all 3 marks awarded |
| $-80$ | A1 | |
---\begin{enumerate}
\item A sequence of numbers $a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots$ is defined by
\end{enumerate}
$$a _ { n + 1 } = \frac { k \left( a _ { n } + 2 \right) } { a _ { n } } \quad n \in \mathbb { N }$$
where $k$ is a constant.\\
Given that
\begin{itemize}
\item the sequence is a periodic sequence of order 3
\item $a _ { 1 } = 2$\\
(a) show that
\end{itemize}
$$k ^ { 2 } + k - 2 = 0$$
(b) For this sequence explain why $k \neq 1$\\
(c) Find the value of
$$\sum _ { r = 1 } ^ { 80 } a _ { r }$$
\hfill \mbox{\textit{Edexcel Paper 1 2020 Q13 [7]}}