Objects in a PyramidDate: 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/ |
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