Square pyramid

• Imagine a square pyramid, with a base of area a---a, of decreasing area up to a point at the top. There are n layers of height h/n. A pyramid of an increasing number of increasingly thinner layers can be constructed so that the height stays a constant total of h.
  
  The volume of such a stack of squares is (from largest to smallest) the sum of area times height of each level: a---a---(h/n) + [(n-1)a/n]---[(n-1)a/n]---(h/n) + [(n-2)a/n]---[(n-2)a/n]---(h/n) + ... + [a/n]---[a/n]---(h/n) = a---a---h --- 1/n --- [1 + (n-1)^2/n^2 + (n-2)^2/n^2 + ... + (1/n)(1/n)]
  
  Note that this can be written as a---a---h --- 1/n^3 --- ['i^2], where the summation equals n(n+1)(2n+1)/6, which can be proven by mathematical induction. If the levels get thinner and thinner, then n goes to infinity. The primary term of n's is (2n^3)/(6n^3), which goes to 1/3 as n gets large.
  
  So the volume of the pyramid is a---a---h/3.
  
  

Mathematical induction

• That the sum of squares 1^2 + 2^2 + ... + n^2 equals n(n+1)(2n+1)/6 can be proven by mathematical induction, that is, prove that the formula holds for n=1. Then show that IF it holds for n, THEN it holds for n+1. Therefore it holds for all positive integers n.
  
  For n=1, that the equality holds is trivial.
  
  Now suppose it holds for n. Then 1^2 + 2^2 + ... + n^2 + (n+1)^2 equals
  n(n+1)(2n+1)/6 + (n+1). The aim is to rearrange this into the form (n+1)((n+1)+1)(2(n+1)+1)/6. That will show that if the sum of squares up to n^2 equals, n(n+1)(2n+1)/6, then the same equality holds for n+1.
  
  n(n+1)(2n+1)/6 + (n+1) = (2n^3+2n^2+n^2+n)/6 +n^2+2n+1 = [(2n^3+3n^2+n)+(6n^2+12n+6)]/6 = (2n^3+6n^2+4n+3n^2+9n+6)/6 = (n^2+3n+2)(2n+3)/6 = (n+1)(n+2)(2(n+1)+1)/6
  which was to be proven.
  
  

A generalization

• Note that the volume of a square pyramid is 1/3 the volume of a block of height h with a square base. The shape of the base didn't affect this result. It is therefore a generalizable result.
  
  It can be shown that a figure with the same shape for each level, converging to a point at the top, is 1/3 the volume of a figure of the same height and constant shape size at all heights.
  
  Therefore, note that the formula for a triangular block of sides a and height h is ha^2sqrt(3)/4. A triangular pyramid of sides a and height h is therefore ha^2sqrt(3)/12.
  
  Also, the volume of a cylinder is pi --- radius^2 --- height. The volume of a cone is therefore pi --- radius^2 --- height/3.