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Geometric representation of the square pyramidal number 1+4+9+16=30.

In mathematics, a pyramid number, or square pyramidal number, is a figurate number that represents a pyramid with a base and four sides. These numbers can be expressed in a formula as

$\sum_{k=1}^nk^2={n(n + 1)(2n + 1) \over 6}={2n^3 + 3n^2 + n \over 6}$

that is, by adding up the squares of the first n integers, or by multiplying the nth pronic number by the nth odd number. By mathematical induction it is possible to derive one formula from the other. An equivalent formula is given in Fibonacci's Liber Abaci (1202, ch. II.12).

This is a special case of Faulhaber's formula.

The first few pyramid numbers are:

1, 5, 14, 30, 55, 91, 140, 204, 285, 385, 506, 650, 819

(sequence A000330 in OEIS).

Pyramid numbers can be modelled in physical space with a given number of balls and a square frame that hold in place the number of balls forming the base, that is, n². They also solve the problem of counting the number of squares in an n × n grid.

[edit] Proof for the formula of the sum of squares

[edit] First Proof

The first proof can be given by induction

$P(n): S(n)=1^2+2^2+...n^2=\frac{n(n+1)(2n+1)}{6}$

P(1) is true trivially

Let P(n) be true from some n

S (n + 1) = S (n) + (n + 1) 2

$S(n+1) = \frac{n(n+1)(2n+1)}{6}+(n+1)^2$

$S(n+1) = \frac{(n+1)(n+2)(2n+3)}{6}$

Hence $P(n) \Rightarrow P(n+1)$

Thus by Principle of mathematical induction, We have proved the claim

[edit] Second Proof

(k + 1) 3 = k 3 + 3 k 2 + 3 k + 1

(k + 1) 3 - k 3 = 3 k 2 + 3 k + 1

$\sum_{k=1}^{n} ((k+1)^3-k^3) = \sum_{k=1}^{n} (3k^2+3k+1)$

$(n+1)^3 -1 = 3 \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} 3k + \sum_{k=1}^{n}1$ (On the LHS, all the terms will get cancelled (Except (n+1) and 1))

$\sum_{k=1}^{n} k^2 = \frac{(n+1)^3 -\sum_{k=1}^{n} 3k - \sum_{k=1}^{n} 1 -1 }{3}$

Hence, : $S(n)=1^2+2^2+...n^2=\frac{n(n+1)(2n+1)}{6}$

[edit] Third Proof

The equivalence between the sum of squares and the cubic polynomial may also be shown by a double counting proof in which one counts in two different ways the number of ways to choose three numbers x, y, and z from the set {1, 2, 3, ... n + 1}, in such a way that z > x and z > y.

First, we fix z and consider the number of ways of choosing x and y. If z = 1 then there are no values of x and y that satisfy the inequality, if z = 2 then the only possible choice is x = y = 1, if z = 3 then x and y may independently be chosen to be either 1 or 2, and in general once z is chosen there are (z − 1)² ways of choosing x and y. Hence the total number of ways of choosing x, y, and z is $1^2+2^2+\cdots+n^2$ .

Another way to count the same set of triples of numbers is to classify them according to which of the three numbers are distinct from each other. The number of ways of choosing three distinct numbers from this set is the binomial coefficient $\scriptstyle\binom{n+1}{3}$ , and each triple of distinct numbers gives us two choices of x, y, and z: the largest of the three numbers must be z while the other two numbers can be assigned to x and y in either of two ways. Thus the number of triples of this type is $\scriptstyle2\binom{n+1}{3}$ . The number of ways of choosing two distinct numbers from this set is the binomial coefficient $\scriptstyle\binom{n+1}{2}$ ; for each such pair, the larger of the two numbers in the pair must be z and x and y must both be equal to the smaller of the two. Therefore, there are $\scriptstyle\binom{n+1}{2}$ triples x, y, and z that come from a pair of distinct numbers in this way. The total number of triples of both types is

$1^2+2^2+\cdots+n^2 = 2\binom{n+1}{3}+\binom{n+1}{2} = \frac{n(n+1)(2n+1)}{6}.$

[edit] Relations to other figurate numbers

The pyramid numbers can also be expressed as sums of binomial coefficients (or of two consecutive tetrahedral numbers) thus:

${{n + 2} \choose 3} + {{n + 1} \choose 3}$

in the same way as square numbers are the sums of two consecutive triangular numbers.

We can derive another relation between square pyramidal numbers and tetrahedral numbers: if we let $P n$ be the nth square pyramid number then

$P_n={n(n+1)(2n + 1) \over 6}={2n(2n+2)(2n+1)\over {4*6}}=(1/4)T_{2n}$

where $T n$ is the nth tetrahedral number.

The sum of two consecutive square pyramidal numbers is an octahedral number.

Besides 1, there is only one other number that is both a square and a pyramid number, 4900, the 70th square number and the 24th square pyramidal number. This fact was proven by G. N. Watson in 1918.

[edit] Squares in a square

A common mathematical puzzle involves finding the number of squares in a large n by n square grid. This number can be derived as follows:

The number of 1×1 boxes found in the grid is $n 2$ .
The number of 2×2 boxes found in the grid is $(n - 1) 2$ . These can be counted by counting all of the possible upper-left corners of 2×2 boxes.
The number of k×k boxes (1 ≤ k ≤ n) found in the grid is $(n - k + 1) 2$ . These can be counted by counting all of the possible upper-left corners of k×k boxes.

It follows that the number of squares in an n by n square grid is:

$x = n^2 + (n-1)^2 + (n-2)^2 + (n-3)^2 + \ldots + 1^2$

or:

x = n (n + 1)(2 n + 1) / 6

That is, the solution to the puzzle is given by the square pyramidal numbers.

[edit] See also

[edit] References

Abramowitz, M.; Stegun, I. A. (Eds.) (1964). Handbook of Mathematical Functions. National Bureau of Standards, Applied Math. Series 55. pp. 813. ISBN 0486612724.

Beiler, A. H. (1964). Recreations in the Theory of Numbers. Dover. pp. 194. ISBN 0486210960.

Goldoni, G. (2002). "A visual proof for the sum of the first n squares and for the sum of the first n factorials of order two". 24. The Mathematical Intelligencer. pp. 67–69.

Sigler, Laurence E. (trans.) (2002). Fibonacci's Liber Abaci. Springer-Verlag. pp. 260–261. ISBN 0-387-95419-8.

[edit] External links

Weisstein, Eric W., "Square Pyramidal Number" from MathWorld.

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