TMUA: Sequences and recurrence relations

Sequences given by a formula for the \(n\)th term or by a recurrence relation \(x_{n+1}=f(x_n)\).

Section 1 Part 1 · MM2 Sequences and series · spec MM2.1
Papers 1 and 2
7 practice questions
No calculator

Practise sequences and recurrence relations →Timed TMUA paper

What the specification covers

nth-term formulas; x(n+1) = f(x(n))
Periodic sequences: find the cycle, then use remainders
Increasing / decreasing / convergent behaviour

Key ideas

Work out the first few terms exactly. Many TMUA recurrences are periodic: once a term repeats, the pattern repeats.
For period \(p\), \(x_n=x_r\) where \(r\) is the remainder of \(n\) on division by \(p\) (use \(r=p\) for remainder 0).
A sequence is increasing if \(x_{n+1}>x_n\) for all \(n\); check the sign of \(x_{n+1}-x_n\).

Common mistakes

Off-by-one errors: if \(x_1\) is the first term, \(x_{n}\) with \(n\equiv1\pmod 3\) equals \(x_1\).
Keep fractions exact; decimals hide a returning value.

Exam tip

Always compute until you see a repeat before trying anything clever.

Worked example

A sequence is defined by \(x_1=2\) and \(x_{n+1}=\dfrac{1}{1-x_n}\) for \(n\ge1\). What is \(x_{2026}\)?

\(2\)
\(\frac{1}{2}\)
\(-1\)
\(1\)
\(-2\)

Answer: A: \(2\)

\(x_2=\tfrac{1}{1-2}=-1\), \(x_3=\tfrac{1}{2}\), \(x_4=\tfrac{1}{1/2}=2=x_1\). The sequence repeats with period 3. \(2026=3\times675+1\), so \(x_{2026}=x_1=2\).

Practice questions

Try each question before opening the solution. No calculator: the TMUA does not allow one. Every answer here was re-checked independently by computer before it was published.

Question 1

A sequence is defined by \(x_1=5\) and \(x_{n+1}=\dfrac{x_n+6}{x_n-1}\) for \(n\ge1\). What is \(x_{2027}\)?

\(-5\)
\(\frac{11}{4}\)
\(\frac{1}{5}\)
\(11\)
\(5\)

Show the answer and solution

Answer: E: \(5\)

\(x_2=\dfrac{5+6}{5-1}=\frac{11}{4}\), and \(x_3=\dfrac{x_2+6}{x_2-1}=5=x_1\) (in general \(f(f(x))=x\) for this map). So the sequence alternates, and \(x_{2027}=x_1\) \(5\).

Question 2

A sequence is defined by \(x_1=4\) and \(x_{n+1}=\dfrac{x_n+6}{x_n-1}\) for \(n\ge1\). What is \(x_{100}\)?

\(-4\)
\(10\)
\(4\)
\(\frac{10}{3}\)
\(\frac{1}{4}\)

Show the answer and solution

Answer: D: \(\frac{10}{3}\)

\(x_2=\dfrac{4+6}{4-1}=\frac{10}{3}\), and \(x_3=\dfrac{x_2+6}{x_2-1}=4=x_1\) (in general \(f(f(x))=x\) for this map). So the sequence alternates, and \(x_{100}=x_2\) \(\frac{10}{3}\).

Question 3

A sequence is defined by \(x_1=3\) and \(x_{n+1}=\dfrac{x_n+7}{x_n-1}\) for \(n\ge1\). What is \(x_{101}\)?

\(10\)
\(5\)
\(\frac{1}{3}\)
\(3\)
\(-3\)

Show the answer and solution

Answer: D: \(3\)

\(x_2=\dfrac{3+7}{3-1}=5\), and \(x_3=\dfrac{x_2+7}{x_2-1}=3=x_1\) (in general \(f(f(x))=x\) for this map). So the sequence alternates, and \(x_{101}=x_1\) \(3\).

Question 4

A sequence is defined by \(x_1=4\) and \(x_{n+1}=\dfrac{x_n+2}{x_n-1}\) for \(n\ge1\). What is \(x_{100}\)?

\(4\)
\(\frac{1}{4}\)
\(6\)
\(2\)
\(-4\)

Show the answer and solution

Answer: D: \(2\)

\(x_2=\dfrac{4+2}{4-1}=2\), and \(x_3=\dfrac{x_2+2}{x_2-1}=4=x_1\) (in general \(f(f(x))=x\) for this map). So the sequence alternates, and \(x_{100}=x_2\) \(2\).

Question 5

A sequence is defined by \(x_1=4\) and \(x_{n+1}=\dfrac{x_n+2}{x_n-1}\) for \(n\ge1\). What is \(x_{101}\)?

\(2\)
\(4\)
\(\frac{1}{4}\)
\(-4\)
\(6\)

Show the answer and solution

Answer: B: \(4\)

\(x_2=\dfrac{4+2}{4-1}=2\), and \(x_3=\dfrac{x_2+2}{x_2-1}=4=x_1\) (in general \(f(f(x))=x\) for this map). So the sequence alternates, and \(x_{101}=x_1\) \(4\).

Question 6

A sequence is defined by \(x_1=4\) and \(x_{n+1}=\dfrac{x_n+2}{x_n-1}\) for \(n\ge1\). What is \(x_{2026}\)?

\(4\)
\(6\)
\(\frac{1}{4}\)
\(2\)
\(-4\)

Show the answer and solution

Answer: D: \(2\)

\(x_2=\dfrac{4+2}{4-1}=2\), and \(x_3=\dfrac{x_2+2}{x_2-1}=4=x_1\) (in general \(f(f(x))=x\) for this map). So the sequence alternates, and \(x_{2026}=x_2\) \(2\).

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