Edexcel C1 2014 June — Question 3 5 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Year2014
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicArithmetic Sequences and Series
TypeRecurrence relation: find parameter from given term
DifficultyModerate -0.5 This is a straightforward recurrence relation question requiring simple substitution to find a₂ = 4k - 3, then a₃ = 4(4k-3) - 3, followed by solving a linear equation a₁ + a₂ + a₃ = 66. It's slightly easier than average as it involves only basic algebraic manipulation with no conceptual challenges beyond understanding the recurrence notation.
Spec1.04e Sequences: nth term and recurrence relations1.04g Sigma notation: for sums of series

A sequence \(a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots\) is defined by $$\begin{array} { l l } a _ { n + 1 } = 4 a _ { n } - 3 , & n \geqslant 1 \\ a _ { 1 } = k , & \text { where } k \text { is a positive integer. } \end{array}$$
  1. Write down an expression for \(a _ { 2 }\) in terms of \(k\). Given that \(\sum _ { r = 1 } ^ { 3 } a _ { r } = 66\)
  2. find the value of \(k\).

Question 3:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\((a_2 =)\ 4k-3\)B1 \(4k-3\) cao
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(a_3 = 4(4k-3)-3\)M1 An attempt to find \(a_3\) from iterative formula \(a_3=4a_2-3\). Condone bracketing errors
\(\sum_{r=1}^{3}a_r = k+4k-3+4(4k-3)-3 = ..k\pm...\)M1 Attempt to sum \(a_1, a_2\) and \(a_3\) to get a linear expression in \(k\). (Sum of Arithmetic series is M0)
\(21k-18=66 \Rightarrow k=...\)dM1 Sets their linear expression to 66 and solves for \(k\). Dependent on previous M mark
\(k=4\)A1 cao \(k=4\)
## Question 3:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(a_2 =)\ 4k-3$ | B1 | $4k-3$ cao |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $a_3 = 4(4k-3)-3$ | M1 | An attempt to find $a_3$ from iterative formula $a_3=4a_2-3$. Condone bracketing errors |
| $\sum_{r=1}^{3}a_r = k+4k-3+4(4k-3)-3 = ..k\pm...$ | M1 | Attempt to sum $a_1, a_2$ and $a_3$ to get a linear expression in $k$. (Sum of Arithmetic series is M0) |
| $21k-18=66 \Rightarrow k=...$ | dM1 | Sets their linear expression to 66 and solves for $k$. Dependent on previous M mark |
| $k=4$ | A1 | cao $k=4$ |

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

$$\begin{array} { l l } 
a _ { n + 1 } = 4 a _ { n } - 3 , & n \geqslant 1 \\
a _ { 1 } = k , & \text { where } k \text { is a positive integer. }
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item Write down an expression for $a _ { 2 }$ in terms of $k$.

Given that $\sum _ { r = 1 } ^ { 3 } a _ { r } = 66$
\item find the value of $k$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C1 2014 Q3 [5]}}