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.

Posted by Nathan Hirtz on 2004-12-07 07:53:30