Consider the numerical pyramid below, formed by simply putting down the series of odd numbers into a pyramid.
1
3 5
7 9 11
13 15 17 19
. . .
Find a formula for the sum of the numbers in the nth row, and prove it.
In a series of m odd numbers the last number is 2m - 1 and the sum of the numbers is (2m - 1 + 1)m/2 = m^2
In a pyramid of odd numbers of n rows there are (n + 1)n/2 numbers. The sum of the numbers is (n + 1)^2(n^2)/4 = (n^4 + 2n^3 + n^2)/4.
In a pyramid of odd numbers of (n - 1) rows there are n(n - 1)/2 numbers. The sum of the numbers is n^2(n - 1)^2/4 = (n^4 - 2n^3 + n^2)/4.
Therefore, the sum of the numbers in the nth row is 4n^3/4 = n^3