# How do you find the general term for a sequence?

Jul 18, 2015

It depends.

#### Explanation:

There are many types of sequences. Some of the interesting ones can be found at the online encyclopedia of integer sequences at https://oeis.org/

Let's look at some simple types:

Arithmetic Sequences

${a}_{n} = {a}_{0} + \mathrm{dn}$

e.g. $2 , 4 , 6 , 8 , \ldots$

There is a common difference between each pair of terms.

If you find a common difference between each pair of terms, then you can determine ${a}_{0}$ and $d$, then use the general formula for arithmetic sequences.

Geometric Sequences

${a}_{n} = {a}_{0} \cdot {r}^{n}$

e.g. $2 , 4 , 8 , 16 , \ldots$

There is a common ratio between each pair of terms.

If you find a common ratio between pairs of terms, then you have a geometric sequence and you should be able to determine ${a}_{0}$ and $r$ so that you can use the general formula for terms of a geometric sequence.

Iterative Sequences

After the initial term or two, the following terms are defined in terms of the preceding ones.

e.g. Fibonacci

${a}_{0} = 0$
${a}_{1} = 1$
${a}_{n + 2} = {a}_{n} + {a}_{n + 1}$

For this sequence we find: ${a}_{n} = \frac{{\phi}^{n} - {\left(- \phi\right)}^{- n}}{\sqrt{5}}$ where $\phi = \frac{1 + \sqrt{5}}{2}$

There are many ways to make these iterative rules, so there is no universal method to provide an expression for ${a}_{n}$

Polynomial Sequences

If the terms of a sequence are given by a polynomial, then given the first few terms of the sequence you can find the polynomial.

e.g.

$\textcolor{red}{1} , 2 , 4 , 7 , 11 , \ldots$

Form the sequence of differences of these values:

$\textcolor{red}{1} , 2 , 3 , 4 , \ldots$

Form the sequence of differences of these values:

$\textcolor{red}{1} , 1 , 1 , \ldots$

Once you reach a constant sequence like this, pick out the initial terms from each sequence. In this case $1$, $1$ and $1$.

These form the coefficients of a polynomial expression:

a_n = color(red)(1)/(0!) + (color(red)(1)*n)/(1!) + (color(red)(1)*n(n-1))/(2!)

$= {n}^{2} / 2 + \frac{n}{2} + 1$