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Triangular Numbers: Find n As a Function of s

Date: 10/13/2002 at 17:04:04
From: Hardev
Subject: Sequences

s = 2, 3, 4, 5

n = 1, 4, 10, 20 

I need to find the relation between s and n, where n is the subject, 
i.e. n = ....

I have tried to answer by spotting patterns between s, 2s, 3s, etc., 
but am not getting anywhere.

I have noticed that the difference between each n is the triangular 
numbers, which have the general formula: (s/2)(s+1) but still can't 
find the relation.

Could you please help me?

Date: 10/14/2002 at 02:04:18
From: Doctor Greenie
Subject: Re: Sequences

Hello, Hardev -

One way to find the formula for n as a function of s is to use the 
method of finite differences. Here is a link to a page in the Dr. 
Math archives where this method is described, with examples:

   Method of Finite Differences 

You can try using this method if you want. If you haven't seen it 
before, it might be a good exercise to go through. The method of 
finite differences can be a very useful tool to have in your bag of 
tricks; using it becomes relatively easy with a little practice.

However, in this case, your observation about the differences being 
triangular numbers leads to a different possible plan of attack for 
this problem that is probably much less work than using the method of 
finite differences.

As you might know, the triangular numbers appear in Pascal's triangle; 
this means your sequence of numbers probably appears in Pascal's 
triangle also. Let's see if we can find it and use our knowledge of 
Pascal's triangle to determine the formula for your sequence.

                 1   1
               1   2   1
             1   3   3   1
           1   4   6   4   1
         1   5  10  10   5   1
       1   6  15  20  15   6   1
     1   7  21  35  35  21   7   1

The numbers in your sequence - 1, 4, 10, 20, ... - are in the 4th 
diagonal row of Pascal's triangle. Given the fact that the j-th entry 
in the i-th row of Pascal's triangle is C(i,j), we have the following:

   n(2) = C(3,0)
   n(3) = C(4,1)
   n(4) = C(5,2)
   n(5) = C(6,3)

From this pattern we can conclude that

                       (s+1)(s)(s-1)   s(s^2-1)
   n(s) = C(s+1,s-2) = ------------- = --------
                         (3)(2)(1)         6

I hope this helps.  Please write back if you have any further 
questions about any of this.

- Doctor Greenie, The Math Forum 

Date: 10/14/2002 at 13:03:21
From: Hardev
Subject: Thank you (Sequences)

Hello Doctor Greenie, 

Thank you very much for replying to my question. I had been stuck on 
the same part of my mathematics investigation for four days, getting 
extremely frustrated. Thanks to you I have been able to take my 
investigation as far as it would go.

Thank you again.
Associated Topics:
High School Sequences, Series

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