Edexcel P2 2019 June — Question 1 4 marks

Exam BoardEdexcel
ModuleP2 (Pure Mathematics 2)
Year2019
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeSimple recurrence evaluation
DifficultyEasy -1.3 This is a straightforward recurrence relation question requiring only direct substitution to find the pattern (alternating between 3 and 1), then arithmetic to evaluate the sum. No problem-solving insight needed—purely mechanical application of the recurrence formula and recognition of periodicity.
Spec1.04e Sequences: nth term and recurrence relations1.04g Sigma notation: for sums of series

  1. A sequence \(a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots\) is defined by
$$\begin{aligned} a _ { n + 1 } & = 4 - a _ { n } \\ a _ { 1 } & = 3 \end{aligned}$$ Find the value of
    1. \(a _ { 2 }\)
    2. \(a _ { 107 }\)
  1. \(\sum _ { n = 1 } ^ { 200 } \left( 2 a _ { n } - 1 \right)\)

AnswerMarks Guidance
(a)(i) \(a_2 = 1\)B1 Accept the sight of 1. Ignore incorrect working
(a)(ii) \(a_{107} = 3\)B1 Accept sight of just 3. Ignore incorrect working. If there are lots of 1's and 3's without reference to any suffices they need to choose 3.
(b) \(\sum_{n=1}^{200}(2a_n - 1) = 5 + 1 + 5 + 1 + ...... + 5 + 1 = 100 \times (5+1) = 600\)M1, A1 M1: Establishes an attempt to find the sum of a series with two distinct terms. Look for \(100 \times a + 100 \times b\) or \(200 \times a + 200 \times b\) where \(a\) and \(b\) are allowable terms. Examples of allowable terms are \(a, b = 1,5\) (which are correct); \(a, b = 1,3\) (which are the values for (a)); \(a, b = 3,7\) (which is using \(2a_n + 1\)); \(a, b = 0,5\) (which is a slip on the first value). Methods using AP (and GP) formulae are common and score 0 marks. A1: 600. 600 should be awarded both marks as long as no incorrect working is seen
**(a)(i)** $a_2 = 1$ | B1 | Accept the sight of 1. Ignore incorrect working

**(a)(ii)** $a_{107} = 3$ | B1 | Accept sight of just 3. Ignore incorrect working. If there are lots of 1's and 3's without reference to any suffices they need to choose 3.

**(b)** $\sum_{n=1}^{200}(2a_n - 1) = 5 + 1 + 5 + 1 + ...... + 5 + 1 = 100 \times (5+1) = 600$ | M1, A1 | M1: Establishes an attempt to find the sum of a series with two distinct terms. Look for $100 \times a + 100 \times b$ or $200 \times a + 200 \times b$ where $a$ and $b$ are allowable terms. Examples of allowable terms are $a, b = 1,5$ (which are correct); $a, b = 1,3$ (which are the values for (a)); $a, b = 3,7$ (which is using $2a_n + 1$); $a, b = 0,5$ (which is a slip on the first value). Methods using AP (and GP) formulae are common and score 0 marks. A1: 600. 600 should be awarded both marks as long as no incorrect working is seen | (4 marks)

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\begin{enumerate}
  \item A sequence $a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots$ is defined by
\end{enumerate}

$$\begin{aligned}
a _ { n + 1 } & = 4 - a _ { n } \\
a _ { 1 } & = 3
\end{aligned}$$

Find the value of\\
(a) (i) $a _ { 2 }$\\
(ii) $a _ { 107 }$\\
(b) $\sum _ { n = 1 } ^ { 200 } \left( 2 a _ { n } - 1 \right)$\\

\hfill \mbox{\textit{Edexcel P2 2019 Q1 [4]}}