Sequences and recurrence relationsEdexcel International A Level Maths: Revision notes

A sequence is an ordered list of numbers $u_1,u_2,u_3,\dots$, where $u_n$ is the $n$th term. A sequence can be defined in two ways:

• by a formula for the $n$th term, such as $u_n=n^2-6n+10$, which lets you find any term directly: $u_5=25-30+10=5$;
• by a recurrence relation, which gives each term from the one before it.

A recurrence relation of the form $u_{n+1}=f(u_n)$ gives each term from the previous one, so you also need a starting value such as $u_1$. For $u_{n+1}=2u_n+3$ with $u_1=5$: $u_2=13$, $u_3=29$, $u_4=61$. Unknown constants are found by forming equations from known terms. If $u_{n+1}=pu_n+q$ with $u_1=2$, $u_2=7$, $u_3=22$, then $7=2p+q$ and $22=7p+q$, so $p=3$, $q=1$. Work term by term and keep each value exact until the end.

A sequence is increasing if $u_{n+1}>u_n$ for all $n$, and decreasing if $u_{n+1}<u_n$ for all $n$. To test, find $u_{n+1}-u_n$ and decide its sign. For $u_n=n^2-6n+10$: $u_{n+1}-u_n=2n-5$, which is negative for $n\le2$ and positive for $n\ge3$. The terms $5,2,1,2,5,10$ fall and then rise, so the sequence is neither increasing nor decreasing overall. For a recurrence such as $u_{n+1}=3u_n+1$ with $u_1=2$: $u_{n+1}-u_n=2u_n+1>0$ because every term is positive, so the sequence is increasing.

A sequence is periodic if its terms repeat in a cycle: $u_{n+k}=u_n$ for all $n$. The smallest such $k$ is the order of the sequence. Example: $u_1=2$, $u_{n+1}=\frac{1}{1-u_n}$ gives $2,-1,\frac12,2,-1,\frac12,\dots$ with order $3$. To find $u_{100}$, divide by the order: $100=3\times33+1$, so $u_{100}=u_1=2$. Formulas with $(-1)^n$ are periodic too: $u_n=6+3(-1)^n$ alternates $3,9,3,9,\dots$ and has order $2$.

For sequence questions: write the first few terms to see the pattern; substitute carefully into the formula or recurrence; use $u_{n+1}-u_n$ to describe the sequence; and give the order for periodic sequences. Check by substituting back: a value of $k$ or $p$ you find should reproduce the given terms.