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Objects in a Pyramid


Date: 7/8/96 at 9:6:16
From: Anonymous
Subject: Objects in a pyramid

Dear Dr. Math,

A number of objects are stacked in a triangular pyramid (e.g., top 
layer 1, next layer 3, next layer 6, etc.).  How many objects are in 
the nth layer from the top?  What is the total number of objects in 
the top n layers?  I need formulas to prove these.

Thanks for your help.

Nancy Geldermann


Date: 7/8/96 at 16:56:5
From: Doctor Pete
Subject: Re: Objects in a pyramid

Consider a single layer (shown below is the 5th layer):

                *
               * *
              * * *
             * * * *
            * * * * * 

Clearly, the n(th) layer contains 1+2+...+n objects.  Let this value 
be S[n]. Note:

      S[n] =   1 +   2 +   3 + ... + n-2 + n-1 +   n
    + S[n] =   n + n-1 + n-2 + ... +   3 +   2 +   1
    -------------------------------------------------
     2S[n] = n+1 + n+1 + n+1 + ... + n+1 + n+1 + n+1
           = n(n+1)

   => S[n] = n(n+1)/2 .

Now, say we want to know the number of objects in the top m layers.  
We have to take the sum of all these S[n] from n=1 to m.  That is, we 
must find S[1]+S[2]+...+S[m].  Call this T[m].  Writing this in sigma 
notation,

              m                    m          
            \---\  n(n+1)     1  \---\        
     T[m] =  >    -------- = ---  >    n^2 + n
            /---/    2        2  /---/        
             n=1                  n=1         

                  m               m             m
             1  \---\        1  \---\      1  \---\       m(m+1)
          = ---  >    n^2 + ---  >    n = ---  >    n^2 + ------
             2  /---/        2  /---/      2  /---/         4
                 n=1             n=1           n=1

Now we must find what the sum 1^2 + 2^2 + 3^2 + ... + m^2 is.  I 
claim that this is m(m+1)(2m+1)/6.  We'll prove this by induction.  
For m=1, this is true, since 1(1+1)(2+1)/6 = 1 = 1^2.  Suppose this 
is true for some value of m.  Then 1^2 + 2^2 + 3^2 + ... + m^2 + 
(m+1)^2 = m(m+1)(2m+1)/6 + (m+1)^2 = (m+1)((m(2m+1)/6)+(m+1)) = 
(m+1)(2m^2+m+6m+6)/6 = (m+1)(2m^2+7m+6)/6 = (m+1)(m+2)(2m+3)/6 = 
(m+1)(m+1+1)(2(m+1)+1)/6, so our assertion is true for m+1. 
Hence the sum of the first m squares is m(m+1)(2m+1)/6.  Therefore,

             m(m+1)(2m+1)   m(m+1)   m(m+1)            m(m+1)(m+2)
     T[m] =  ------------ + ------ = ------ (2m+1+3) = ----------- .
                  12          4        12                   6

-Doctor Pete,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Basic Algebra
High School Discrete Mathematics

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