Edexcel C1 2013 June — Question 6 9 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Year2013
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicArithmetic Sequences and Series
TypeRecurrence relation: evaluate sum
DifficultyModerate -0.5 This is a straightforward recurrence relation question requiring systematic substitution and basic algebraic manipulation. Parts (a)-(c) involve routine substitution into the given formula, while part (d) requires recognizing that the sequence becomes periodic (x₃=x₁=1 implies the sequence repeats). The algebraic manipulation is standard C1 level, and the pattern recognition in part (d), while requiring some insight, is guided by the structure of the question. Slightly easier than average due to the step-by-step scaffolding.
Spec1.04e Sequences: nth term and recurrence relations1.04g Sigma notation: for sums of series

6. A sequence \(x _ { 1 } , x _ { 2 } , x _ { 3 } \ldots\) is defined by $$\begin{gathered} x _ { 1 } = 1 \\ x _ { n + 1 } = \left( x _ { n } \right) ^ { 2 } - k x _ { n } , \quad n \geqslant 1 \end{gathered}$$ where \(k\) is a constant, \(k \neq 0\)
  1. Find an expression for \(x _ { 2 }\) in terms of \(k\).
  2. Show that \(x _ { 3 } = 1 - 3 k + 2 k ^ { 2 }\) Given also that \(x _ { 3 } = 1\),
  3. calculate the value of \(k\).
  4. Hence find the value of \(\sum _ { n = 1 } ^ { 100 } x _ { n }\)

Question 6(a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(x_2 = 1-k\)B1 Accept unsimplified e.g. \(1^2 - 1k\)
Question 6(b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(x_3 = (1-k)^2 - k(1-k)\)M1 Attempt to substitute their \(x_2\) into \(x_3 = (x_2)^2 - kx_2\) with their \(x_2\) in terms of \(k\)
\(= 1 - 3k + 2k^2\)A1* Answer given
Question 6(c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(1 - 3k + 2k^2 = 1\)M1 Setting \(1 - 3k + 2k^2 = 1\)
\(k(2k-3) = 0 \Rightarrow k = \ldots\)dM1 Solving quadratic to obtain a value for \(k\). Dependent on previous M1
\(k = \frac{3}{2}\)A1 cao and cso (ignore any reference to \(k=0\))
Question 6(d):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\sum_{n=1}^{100} x_n = 1 + \left(-\frac{1}{2}\right) + 1 + \ldots\) or \(1 + (1-k') + 1 + \ldots\)M1 Writing out at least 3 terms with third term equal to first. Allow in terms of \(k\) as well as numerical values. Evidence that sequence oscillates between 1 and \(1-k\)
\(50 \times \frac{1}{2}\) or \(50\times1 - 50\times\frac{1}{2}\) or \(\frac{1}{2}\times50\times\left(1-\frac{1}{2}\right)\)M1 Attempt to combine terms correctly. Can be in terms of \(k\), e.g. \(100-50k\)
\(= 25\)A1 Allow equivalent fraction e.g. 50/2 or 100/4
## Question 6(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_2 = 1-k$ | B1 | Accept unsimplified e.g. $1^2 - 1k$ |

## Question 6(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_3 = (1-k)^2 - k(1-k)$ | M1 | Attempt to substitute their $x_2$ into $x_3 = (x_2)^2 - kx_2$ with their $x_2$ in terms of $k$ |
| $= 1 - 3k + 2k^2$ | A1* | Answer given |

## Question 6(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $1 - 3k + 2k^2 = 1$ | M1 | Setting $1 - 3k + 2k^2 = 1$ |
| $k(2k-3) = 0 \Rightarrow k = \ldots$ | dM1 | Solving quadratic to obtain a value for $k$. Dependent on previous M1 |
| $k = \frac{3}{2}$ | A1 | cao and cso (ignore any reference to $k=0$) |

## Question 6(d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum_{n=1}^{100} x_n = 1 + \left(-\frac{1}{2}\right) + 1 + \ldots$ or $1 + (1-k') + 1 + \ldots$ | M1 | Writing out at least 3 terms with third term equal to first. Allow in terms of $k$ as well as numerical values. Evidence that sequence oscillates between 1 and $1-k$ |
| $50 \times \frac{1}{2}$ or $50\times1 - 50\times\frac{1}{2}$ or $\frac{1}{2}\times50\times\left(1-\frac{1}{2}\right)$ | M1 | Attempt to combine terms correctly. Can be in terms of $k$, e.g. $100-50k$ |
| $= 25$ | A1 | Allow equivalent fraction e.g. 50/2 or 100/4 |
6. A sequence $x _ { 1 } , x _ { 2 } , x _ { 3 } \ldots$ is defined by

$$\begin{gathered}
x _ { 1 } = 1 \\
x _ { n + 1 } = \left( x _ { n } \right) ^ { 2 } - k x _ { n } , \quad n \geqslant 1
\end{gathered}$$

where $k$ is a constant, $k \neq 0$
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $x _ { 2 }$ in terms of $k$.
\item Show that $x _ { 3 } = 1 - 3 k + 2 k ^ { 2 }$

Given also that $x _ { 3 } = 1$,
\item calculate the value of $k$.
\item Hence find the value of $\sum _ { n = 1 } ^ { 100 } x _ { n }$
\end{enumerate}

\hfill \mbox{\textit{Edexcel C1 2013 Q6 [9]}}