OCR MEI C2 — Question 11 12 marks

Exam BoardOCR MEI
ModuleC2 (Core Mathematics 2)
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicArithmetic Sequences and Series
TypePeriodic or repeating sequence
DifficultyChallenging +1.2 This is a multi-part question on recurrence relations requiring systematic calculation and pattern recognition. Parts (i)-(ii) involve routine substitution and algebraic manipulation. Part (iii) requires verifying a product identity using established results. Part (iv) applies the periodic pattern to sum terms. While it requires careful algebra and understanding of periodicity (a C2-appropriate concept), the steps are guided and methodical rather than requiring novel insight. Slightly above average difficulty due to length and the need to connect multiple parts, but well within reach of a competent C2 student.
Spec1.04e Sequences: nth term and recurrence relations

11 The sequence \(a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots \ldots\) where \(a _ { 1 }\) is a given real number is defined by \(a _ { n + 1 } = 1 - \frac { 1 } { a _ { n } }\).
  1. For the case when \(a _ { 1 } = 2\), find \(a _ { 2 } , a _ { 3 }\) and \(a _ { 4 }\). Describe the behaviour of this sequence
  2. For the case when \(a _ { 1 } = k\), where \(k\) is an integer greater than 1 , find \(a _ { 2 }\) in terms of \(k\) as a single fraction.
    Find also \(a _ { 3 }\) in its simplest form and hence deduce that \(a _ { 4 } = k\).
  3. Show that \(a _ { 2 } a _ { 3 } a _ { 4 } = - 1\) for any integer \(k\).
  4. When \(a _ { 1 } = 2\) evaluate \(\sum _ { i = 1 } ^ { 99 } a _ { i }\).

AnswerMarks Guidance
(i) \(\frac{1}{2}, -1, 2\)B1, B1
PeriodicB1 Total: 3 marks
(ii) \(a_2 = 1 - \frac{1}{k} = \frac{k-1}{k}\)M1, A1 attempt; correct process
\(a_3 = 1 - \frac{k}{k-1} = \frac{-1}{k-1}\)A1 attempt simplification in terms of \(k\)
\(a_4 = 1 - \frac{k-1}{-1} = k\)A1 Total: 4 marks
(iii) \(\frac{k-1}{k} \times \frac{-1}{k-1} \times k = -1\)M1, A1 Total: 2 marks
(iv) \(\sum_{i=1}^{99} a_i = \left(2 + \frac{1}{2} - 1\right) + \left(2 + \frac{1}{2} - 1\right) + \ldots\)M1 Awareness to take in groups of 3
\(= \frac{3}{2} \times 33\)A1 For 33
\(= 49.5\)A1 c.a.o
**(i)** $\frac{1}{2}, -1, 2$ | B1, B1 |
Periodic | B1 | **Total: 3 marks**

**(ii)** $a_2 = 1 - \frac{1}{k} = \frac{k-1}{k}$ | M1, A1 | attempt; correct process
$a_3 = 1 - \frac{k}{k-1} = \frac{-1}{k-1}$ | A1 | attempt simplification in terms of $k$
$a_4 = 1 - \frac{k-1}{-1} = k$ | A1 | **Total: 4 marks**

**(iii)** $\frac{k-1}{k} \times \frac{-1}{k-1} \times k = -1$ | M1, A1 | **Total: 2 marks**

**(iv)** $\sum_{i=1}^{99} a_i = \left(2 + \frac{1}{2} - 1\right) + \left(2 + \frac{1}{2} - 1\right) + \ldots$ | M1 | Awareness to take in groups of 3
$= \frac{3}{2} \times 33$ | A1 | For 33
$= 49.5$ | A1 | c.a.o | **Total: 3 marks**
11 The sequence $a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots \ldots$ where $a _ { 1 }$ is a given real number is defined by $a _ { n + 1 } = 1 - \frac { 1 } { a _ { n } }$.\\
(i) For the case when $a _ { 1 } = 2$, find $a _ { 2 } , a _ { 3 }$ and $a _ { 4 }$. Describe the behaviour of this sequence\\
(ii) For the case when $a _ { 1 } = k$, where $k$ is an integer greater than 1 , find $a _ { 2 }$ in terms of $k$ as a single fraction.\\
Find also $a _ { 3 }$ in its simplest form and hence deduce that $a _ { 4 } = k$.\\
(iii) Show that $a _ { 2 } a _ { 3 } a _ { 4 } = - 1$ for any integer $k$.\\
(iv) When $a _ { 1 } = 2$ evaluate $\sum _ { i = 1 } ^ { 99 } a _ { i }$.

\hfill \mbox{\textit{OCR MEI C2  Q11 [12]}}