Sequences and recurrence relationsAQA A-Level Maths: Revision notes

A sequence is an ordered list of numbers $u_1,u_2,u_3,\dots$. It can be defined by a formula for the $n$th term, such as $u_n=n^2-3n$, which gives any term directly: $u_5=25-15=10$. To find the position of a given value, solve an equation in $n$. For $a_n=\frac{n}{n+2}$, the first term exceeding $0.9$ needs $\frac{n}{n+2}>0.9$, so $n>18$. The first such term is $a_{19}$, because $a_{18}=0.9$ exactly.

A recurrence relation (or iterative formula) of the form $x_{n+1}=f(x_n)$ builds each term from the one before. It needs a starting value, such as $x_1$. Example: $u_1=3$, $u_{n+1}=2u_n-1$ gives $3,5,9,17,33,\dots$ Work term by term, keeping exact values (fractions) when the question does not ask for decimals. Check a proposed formula by substituting it into the relation: if $u_n=2^n+1$ then $2u_n-1=2^{n+1}+1=u_{n+1}$, and $u_1=3$, so the formula fits.

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 prove it, look at the sign of $u_{n+1}-u_n$. For $a_n=\frac{n}{n+2}$: $a_{n+1}-a_n=\frac{(n+1)(n+2)-n(n+3)}{(n+2)(n+3)}=\frac{2}{(n+2)(n+3)}>0,$ so the sequence is increasing. A few terms are not a proof: you must show the inequality for all $n$. Some sequences are neither, such as $1,-1,1,-1,\dots$

A sequence is periodic with order (period) $k$ if $u_{n+k}=u_n$ for all $n$: the same $k$ terms repeat. Take $x_1=2$ and $x_{n+1}=\frac{1}{1-x_n}$: $2,\ -1,\ \tfrac12,\ 2,\ -1,\ \tfrac12,\dots$ This has order $3$. To find a late term, divide the position by the order and use the remainder: $100=3\times33+1$, so $x_{100}=x_1=2$. For a sum, add up whole cycles and then the leftover terms: one cycle sums to $\frac32$, so the first $100$ terms sum to $33\times\frac32+2=51.5$.

Many questions give a relation containing a constant $k$, e.g. $u_{n+1}=ku_n+4$ with $u_1=1$. Express the first few terms in terms of $k$: $u_2=k+4,\qquad u_3=k(k+4)+4=(k+2)^2.$ Then use the extra information. If $u_3=25$ and $k>0$ then $k=3$ and $u_4=79$. If $u_3=u_1$ then $(k+2)^2=1$, so $k=-1$ (giving $1,3,1,3,\dots$, order $2$) or $k=-3$ (giving a constant sequence $1,1,1,\dots$). Check each solution against any condition stated in the question and describe the resulting sequence.