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
  leading coefficient a/2.

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/