No formula for the nth term needed — just tell a sequence what its next term depends on, and let it generate itself.
9 min read4 worked examples
Two ways to define a sequence
So far, sequences have been defined by an explicit formula for the $n$th term, such as $u_n=a+(n-1)d$ — plug in $n$, and the term pops out directly, independent of any other term. A recurrence relation instead defines each term in terms of the term (or terms) before it, together with a starting value:
$u_{n+1} = f(u_n), \qquad u_1 = \text{given}$
To find $u_{10}$ from a recurrence relation, there's no shortcut — every term from $u_1$ up to $u_9$ must be generated first, each one feeding into the next. This makes recurrence relations well suited to modelling processes that evolve step by step, like compound interest, population growth, or iterative approximations.
Reading the notation
$u_{n+1}=f(u_n)$ means "the next term equals some function of the current term." For example, $u_{n+1}=2u_n+1$ with $u_1=3$ generates $u_2=2(3)+1=7$, then $u_3=2(7)+1=15$, then $u_4=2(15)+1=31$, and so on — each term built directly from its predecessor.
Classifying a sequence's behaviour
Once several terms of a sequence are known, its long-run behaviour can be classified:
Increasing, decreasing, or periodic
Increasing: $u_{n+1}>u_n$ for all $n$ — every term is strictly larger than the one before. Decreasing: $u_{n+1}<u_n$ for all $n$ — every term is strictly smaller than the one before. Periodic: the sequence repeats itself with some fixed order $k$, meaning $u_{n+k}=u_n$ for all $n$ — after $k$ steps, the sequence returns to a value it has already visited, and the cycle then repeats indefinitely.
Not every sequence falls neatly into one of these categories — some oscillate without ever exactly repeating, and others increase for a while before decreasing (or vice versa). Where a question asks for a classification, generating enough terms to see a clear, sustained pattern is usually the most reliable approach.
Worked example 1
Generating terms from a recurrence relation
A sequence is defined by $u_{n+1}=3u_n-4$, with $u_1=2$. Find $u_2$, $u_3$, and $u_4$.
MethodSubstitute each term in turn to generate the next
1
$u_2 = 3u_1-4 = 3(2)-4 = 2$.
Substitute $u_1=2$ into the recurrence relation.
2
$u_3 = 3u_2-4 = 3(2)-4 = 2$.
Substitute the newly found $u_2$ back into the same relation.
3
$u_4 = 3u_3-4 = 3(2)-4 = 2$.
Repeat the process for $u_4$.
▪ solved
The key insight
Every term here equals $2$ — this is a constant sequence, a special (degenerate) case of periodic with order $k=1$. It arises because $u_1=2$ happens to be a fixed point of the relation: solving $u=3u-4$ gives $u=2$ exactly, so the sequence never moves once it starts there.
Worked example 2
Classifying a sequence as increasing or decreasing
A sequence is defined by $u_{n+1}=\dfrac12 u_n + 3$, with $u_1=4$. Generate the first 5 terms and determine whether the sequence is increasing, decreasing, or neither.
MethodGenerate terms, then compare consecutive values
Apply the recurrence relation repeatedly, starting from $u_1=4$.
2
The terms are $4, 5, 5.5, 5.75, 5.875,\dots$ — each term is strictly larger than the one before.
List the generated terms in order to compare them directly.
3
Since $u_{n+1}>u_n$ for every pair generated, the sequence is (strictly) increasing.
Matches the definition of an increasing sequence.
▪ solved
Look deeper
The terms are increasing but by ever-smaller amounts ($+1, +0.5, +0.25, +0.125,\dots$) — the sequence is converging towards a limit of $6$ (the fixed point where $u=\tfrac12u+3$), without ever reaching or exceeding it. Increasing and bounded above are not contradictory.
Worked example 3
Identifying a periodic sequence
A sequence is defined by $u_{n+1}=\dfrac{1}{1-u_n}$, with $u_1=2$. Find $u_2$, $u_3$, and $u_4$, and state the order of the periodicity.
MethodGenerate terms until a value repeats, then read off the order
$u_4 = 2 = u_1$, so the sequence returns to its starting value after 3 steps: $u_{n+3}=u_n$ for all $n$.
A repeated value signals the sequence has entered a cycle.
5
The sequence is periodic with order $k=3$.
The cycle length (number of distinct terms before repeating) defines the order.
▪ solved
Examiner's eye
To confirm periodicity rather than assume it from one repeat, it's good practice to generate at least one further term ($u_5$ should equal $u_2=-1$) — a single coincidence isn't a proof, but the recurrence relation guarantees the cycle continues once one full repeat is seen, since each term depends only on the one before.
Worked example 4
Finding an unknown constant using periodicity · Challenge
A sequence is defined by $u_{n+1}=u_n^2-k$, with $u_1=1$, where $k$ is a constant. Given that the sequence is periodic with order 2, find the value of $k$.
MethodUse the periodicity condition to set up and solve an equation in k
1
$u_2 = u_1^2 - k = 1^2-k = 1-k$.
Substitute $u_1=1$ into the recurrence relation.
2
$u_3 = u_2^2 - k = (1-k)^2 - k$.
Substitute $u_2=1-k$ into the same relation.
3
Periodic with order 2 means $u_3=u_1$, so $(1-k)^2-k = 1$.
Applying the definition $u_{n+2}=u_n$ directly to $n=1$.
Expand and simplify into a factorisable quadratic in $k$.
5
$k=0$ or $k=3$. If $k=0$: $u_2=1-0=1=u_1$, so the sequence is constant (order 1, not genuinely order 2). If $k=3$: $u_2=1-3=-2$, and $u_3=(-2)^2-3=1=u_1$ ✓, with $u_2=-2\ne u_1=1$, confirming a genuine order-2 cycle.
Check each root: $k=0$ gives a degenerate (constant) sequence, not true period-2 behaviour.
6
The value of $k$ is $3$.
Only $k=3$ gives a sequence that is periodic with order exactly 2.
▪ solved
Look deeper — order 1 is a hidden trap inside "order 2"
A constant sequence technically satisfies $u_{n+2}=u_n$ (since every term is equal), so it solves the equation set up from periodicity — but it isn't genuinely period-2, since its true (smallest) order is 1. Whenever a periodicity condition produces multiple solutions, check that each one doesn't secretly collapse to a smaller order than the one required.
Common mistake
Assuming a sequence is increasing or decreasing after checking only the first pair of terms ($u_2$ vs $u_1$) — the definition requires $u_{n+1}>u_n$ (or $<$) for all $n$, and some recurrence relations change direction after a few terms. Generate enough terms to be confident of a sustained, unbroken pattern before classifying.
Check yourself
A sequence is defined by $u_{n+1}=\dfrac{6}{u_n}$, with $u_1=3$. Find $u_2$, $u_3$, and $u_4$, and state the order of the periodicity. (Hint: generate terms one at a time until a value repeats.)
Method
1
$u_2 = \dfrac{6}{u_1} = \dfrac{6}{3} = 2$.
Substitute $u_1=3$.
2
$u_3 = \dfrac{6}{u_2} = \dfrac{6}{2} = 3$.
Substitute $u_2=2$; note $u_3=u_1$ already.
3
$u_4 = \dfrac{6}{u_3} = \dfrac{6}{3} = 2 = u_2$, confirming the cycle continues. The sequence is periodic with order $k=2$.