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### Finding Rules for Number Patterns

```Date: 06/05/2009 at 22:26:33
From: Jerold
Subject: finding the algebraic expression

Find the algebraic expression for the following pattern:
3,5,8,12,17,23,30.

I found out that it increases by 1 more than before.  I tried the
different possibilities n^2, 2n, 2^n and trying to add other
digits but failed.  I have tried similar problems but never found the
algebraic expression for this kind yet.

```

```
Date: 06/06/2009 at 02:35:22
From: Doctor Greenie
Subject: Re: finding the algebraic expression

Hi, Jerold --

Here is a link to a page in the Dr. Math archives where you can find
a discussion of a general method for finding an algebraic expression
for a pattern like this:

Method of Finite Differences
http://mathforum.org/library/drmath/view/53223.html

That method is perhaps a bit advanced for a 13-year-old; so let's
look at (what I think is) a simpler process that is related to the
formal finite differences method.

Let's start by looking at the pattern generated by the simple formula

t(n) = n^2

For this sequence, we get

1, 4, 9, 16, 25, 36, ...

The differences between successive terms are

3, 5, 7, 9, 11, ...

In this example, the differences are increasing by 2 each time; and
our formula is simply "x^2".

Now here is the way you can use this example to find the formula for
any similar sequence in which the differences between terms are
increasing by the same amount each time:

If the differences between successive terms of a sequence are
increasing by a constant amount "a" each time, then the formula
for generating the sequence is a quadratic polynomial with

I state that rule without proof....

Let's apply it to your example to find the formula for the sequence.

The differences in your sequence increase by 1 each time, so the
formula for generating the sequence is a quadratic polynomial

an^2 + bn + c

with leading coefficient 1/2; that is, the formula is of the form

(1/2)n^2 + bn + c

The "bn + c" is a linear polynomial.  If we look at the difference
between the value of "(1/2)n^2)" for each value of n and the
corresponding term of the given sequence, we can determine that
linear polynomial part of our formula.  We find

n   n-th term   (1/2)n^2   difference
-------------------------------------
1       3         .5         2.5
2       5        2.0         3.0
3       8        4.5         3.5
4      12        8.0         4.0
5      17       12.5         4.5
...

A bit of examination, or perhaps some formal algebra, shows that the
formula in terms of n for the numbers in the difference column above
is

(1/2)n + 2

And so the complete formula for your sequence is

(1/2)n^2 + (1/2)n + 2

It is easy to verify that this formula generates the given sequence;
perhaps by using the formula in a slightly different form:

(1/2)n^2 + (1/2)n + 2 = (n^2 + n + 4)/2 = (n(n+1)+4)/2

n   (n(n+1)+4)/2
----------------
1    (1*2+4)/2 = 3
2    (2*3+4)/2 = 5
3    (3*4+4)/2 = 8
4    (4*5+4)/2 = 12
5    (5*6+4)/2 = 17
6    (6*7+4)/2 = 23
7    (7*8+4)/2 = 30

(n^2 + n + 4)/2

- Doctor Greenie, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra
High School Sequences, Series
Middle School Algebra

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