Finding a Formula for a Number Pattern

Date: 09/30/2004 at 00:14:28
From: Kelly
Subject: Sequence pattern/factoring

Hi Dr. Math,

I am in a geometry class and we are learning about sequences and how
to find the patterns in numbers.  I was fine up until our teacher
wrote this up on the board:
     
  0, 3, 8, 15, 24, 35
     
She told us that we had to use factoring to find this answer.

What is confusing about this question is I don't know how to get from
the problem to the answer that was given:

  (n+1)(n-1)

How did my teacher get this?

I know from this problem that the whole pattern is that you add by odd
numbers (like to get from 0 to 3 you go by 3, to get from 3 to 8 you
go by 5, 8 to 15 you go by 7, etc).  But I can't see how to get that
other answer.

Any help would be greatly appreciated.  Thanks for your time.

Date: 09/30/2004 at 09:05:23
From: Doctor Peterson
Subject: Re: Sequence pattern/factoring

Hi, Kelly.

The hint says to factor the numbers in the sequence:

  0, 3, 8, 15, 24, 35

  0, 3, 2*2*2, 3*5, 2*2*2*3, 5*7

What if we look for a way to factor each number into TWO factors so 
that they all look similar? Here is what I come up with:

  0, 1*3, 2*4, 3*5, 4*6, 5*7

When I saw this, I just noticed that the pattern was one of 
multiplying a pair of numbers where both increase by one each time, 
with the first number going 1,2,3... and the second going 3,4,5..., 
etc..

One helpful way to organize your thoughts is to make a chart of the
pattern and include the index, or the number of the result in the
pattern.  Then you can see if there is a consistent relationship
between the pattern and the index.  In this case, you can see that the
first number is always one less than the index, and the second is one 
more:

  index  term
  -----  ----
    1    0*2
    2    1*3
    3    2*4
    4    3*5
    5    4*6
    6    5*7
    n    (n-1)*(n+1)

The second term is 2-1 times 2+1, the third is 3-1 times 3+1, and so
on.  Seeing that this pattern applies to all the terms, the formula 
for the sequence is

  a[n] = (n-1)(n+1)

(Of course, any answer is really just a guess--this is a nice formula, 
so we assume it is what was intended, but in real life sequences 
aren't always nice!)

How did the teacher get it?  Probably because she or whoever invented 
the problem STARTED with (n-1)(n+1) and wrote out that sequence! 
Solving problems like this is very different from making them up.  You 
couldn't just write down any random sequence and then figure out "the 
pattern" to it; random numbers don't generally have any pattern.  The 
pattern is there because someone deliberately made it that way.

Now, you seem to have come up with a perfectly good alternative 
answer, using a standard technique of looking for a way to get from 
one term to the next.  The sequence of differences is

  0, 3, 8, 15, 24, 35

    3, 5, 7,  9, 11

and that pattern is clear enough that you know how to find as many 
terms as you want.  That's a fine answer, unless you were asked to  
find an explicit formula, which means you can find any term in the
sequence by calculating it directly and not having to work your way
through the sequence all the way to the desired term.  How could you
work your pattern into an explicit formula?

I happen to know that the sum of consecutive odd numbers is a square:

  1 + 3 = 4
  1 + 3 + 5 = 9
  1 + 3 + 5 + 7 = 16

and so on.  The formula for this is

  1 + 3 + 5 + ... + (2n-1) = n^2

Your sequence is just my sum with 1 removed from the start! So the 
formula is

  a[n] = n^2 - 1

Check that: if we add 1 to each term of your sequence, we get

  0, 3, 8, 15, 24, 35

  1, 4, 9, 16, 25, 36

(In fact, that is another way one could find the answer!)

And this new formula is equivalent to the one you were told, since the 
difference of squares n^2 - 1 factors as (n-1)(n+1).

Interesting, isn't it? There are a lot of ways to solve this one, and 
they all give the same answer.

If you have any further questions, feel free to write back.


- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 09/30/2004 at 21:03:50
From: Kelly
Subject: Sequence pattern/factoring

Hi Dr. Math,

Thanks for answering my question, and the second idea of how this 
problem can be viewed.  It is really interesting to see what you can 
do with some problems like these.