数列

Consider the sequence $5,8,11,14,\dots$. The term-to-term rule is “add 3 to the previous term.” That is handy for continuing the list, but inefficient if you want the 100th term — you would have to add 99 times. The position-to-term rule (the $n$th term) is a formula into which you substitute the position $n$ to get that term directly, and it is the one exams focus on.

A sequence whose terms increase by a constant amount is linear (arithmetic), and its $n$th term has the form $an+b$. Here $a$ is the common difference between consecutive terms, and $b$ is an adjustment (a “zeroth term”) found from $b=(\text{first term})-a$. Example: for $5,8,11,14,\dots$, $a=3$ and $b=5-3=2$, so the $n$th term is $3n+2$. Substituting $n=1$ gives $5$, which matches.

Once you have the $n$th term, any particular term comes from substituting its position. Example: for the $n$th term $3n+2$, the 10th term is $3\times 10+2=32$ and the 50th term is $3\times 50+2=152$ — no need to add 99 times.

To decide whether a number is a term, set the $n$th term equal to it and solve for $n$. If $n$ is a positive integer the number is in the sequence; if $n$ is a fraction or negative it is not. Example: is $100$ in $4n-1$? $4n-1=100\Rightarrow 4n=101\Rightarrow n=25.25$, not an integer, so $100$ is not a term. By contrast, is $98$ in $5n+3$? $5n+3=98\Rightarrow n=19$, an integer, so $98$ is the 19th term.

If the first differences are not constant but the differences of those differences (the second differences) are constant, the sequence is quadratic and its $n$th term contains $n^2$. A key fact: if the second difference is the constant $D$, the coefficient of $n^2$ in the $n$th term is $\frac{D}{2}$. Example: $2,6,12,20,30,\dots$ has first differences $4,6,8,10$ and second differences $2,2,2$. With $D=2$ the $n^2$-coefficient is $1$, and checking shows the $n$th term is $n^2+n=n(n+1)$ (giving $2,6,12$ at $n=1,2,3$).