Perimeter of Pascal's Triangle

Date: 06/04/2001 at 20:43:23
From: Bobby Mcdonald
Subject: Formula for Pascal's triangle

Is there a general formula for finding the perimeter of Pascal's 
triangle using the number of rows?

Formula for Perimeter of Pascal's Triangle

Number of rows   Perimeter sum
---------------------------------------------------
     2                3
     3                7
     4               13
     5               23
     6               41      

I have this information. Using this, what is a general formula for the 
perimeter N? In other words, what would the perimeter of N be?

Date: 06/07/2001 at 11:50:32
From: Doctor Toughy
Subject: Re: Formula for Pascal's triangle

Hi and thanks for writing to Dr. Math.

In our FAQ on Pascal's Triangle, 

  http://mathforum.org/dr.math/faq/faq.pascal.triangle.html   

there is the following sample triangle:

                             1
                           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

I notice that you created a table for yourself that gives the sum of 
all the numbers in the perimeter of the triangle. This was a very good 
move on your part because quite often in mathematics, and as in this 
case, a table can help you detect a pattern of change as a problem 
grows in size.  

In the FAQ I cited above, the author defined the top row as row zero
such that the above diagram shows row zero through row seven.

The reason the pattern wasn't apparent in your table is that you're
dealing with two different patterns. The sum of the two sides 
increases by two with the addition of each row, while the bottom 
doubles in value. The table you create needs to show both patterns 
separately. I've constructed a new table that does that. The bottom 
"1" on each side will be included with the sum of the bottom row, as 
opposed to being added to the sum of the side perimeter


   row    |   sides   |  bottom  |  total sum of perimeter
  ___________________________________________________
    0     |    0      |    1     |       1
    1     |    1      |    2     |       3
    2     |    3      |    4     |       7
    3     |    5      |    8     |      13
    4     |    7      |   16     |      23
    5     |    9      |   32     |      41
    6     |   11      |   64     |      75
    7     |   13      |  128     |     141

If we let n equal the number of the bottom row, then:
 
    The sum of the two sides = 2n - 1 where n > 0

    The sum of the bottom row = 2^n

If you add these together, you have the formula you're looking for. 

Hope this helps.  If you have any more questions about this or any 
other math topics, please write back.

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