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.
I didn't bother to see if anyone else had done it this way, but I came up with this on my own.
The pattern of first numbers is: 1, 3, 7, 13, 21, ... After some experimentation I discovered the formula for the first term of row N is N²-N+1.
Also, row N has N terms in it: N²-N+1, N²-N+3, ... , N²-N+(2N-1)
If we add the terms of row N together, we get: N(N²-N)+1+3+...+2N-1
1+3+...+2N-1 = 1+2+3+...+2N-1 - (2+4+...+2N-2)
1+3+...+2N-1 = ½(2N-1)(2N) - 2(½(N-1)(N))
1+3+...+2N-1 = 2N²-N-N²+N
Thus, the sum of row N is: N³-N²+2N²-N-N²+N = N³
Therefore, the sum of the numbers in the nth row is n³. Q.E.D.