SPS SPS SM 2022 October — Question 9 7 marks

Exam BoardSPS
ModuleSPS SM (SPS SM)
Year2022
SessionOctober
Marks7
TopicSequences and Series
TypePeriodic Sequences
DifficultyChallenging +1.2 This is a multi-part recurrence relation problem requiring algebraic manipulation to find k, then exploiting periodicity to sum terms. Part (a) requires computing a₂ and a₃, then using the period-3 condition (a₄ = a₁) to derive the quadratic—this is moderately challenging but systematic. Parts (b) and (c) are straightforward once k is found. The problem requires careful algebra and understanding of periodic sequences, placing it above average difficulty but not requiring exceptional insight.
Spec1.04e Sequences: nth term and recurrence relations1.04g Sigma notation: for sums of series

A sequence of numbers \(a_1, a_2, a_3, \ldots\) is defined by $$a_{n+1} = \frac{k(a_n + 2)}{a_n}$$, \(n \in \mathbb{N}\) where \(k\) is a constant. Given that
  • the sequence is a periodic sequence of order 3
  • \(a_1 = 2\)
  1. show that $$k^2 + k - 2 = 0$$ [3]
  2. For this sequence explain why \(k \neq 1\) [1]
  3. Find the value of $$\sum_{r=1}^{80} a_r$$ [3]

A sequence of numbers $a_1, a_2, a_3, \ldots$ is defined by
$$a_{n+1} = \frac{k(a_n + 2)}{a_n}$$, $n \in \mathbb{N}$

where $k$ is a constant.

Given that
\begin{itemize}
\item the sequence is a periodic sequence of order 3
\item $a_1 = 2$
\end{itemize}

\begin{enumerate}[label=(\alph*)]
\item show that
$$k^2 + k - 2 = 0$$ [3]

\item For this sequence explain why $k \neq 1$ [1]

\item Find the value of
$$\sum_{r=1}^{80} a_r$$ [3]
\end{enumerate}

\hfill \mbox{\textit{SPS SPS SM 2022 Q9 [7]}}