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.

The sequence 1,3,7,13..... describes the first element of successive rows, while the sequence 1,5,11, 19,... denotes the last element of the successive rows. Both the above sequence correspond to arithmetico geometric series (reference: http://www.artofproblemsolving.com/Wiki/index.php/Arithmetico-geometric_series ).

Let s(n) = 1 + 3 + 7 + 13 + .....+ t(n), and:

s(n) = 1 + 3 + 7 + .... + t(n-1) + t(n)

So, t(n) = 1 + (2+4+6+....+2(n-1))

= 1 + n(n-1) = n^2 - n + 1

Thus, the first element of the nth row is n^2 - n + 1.

In a similar manner, it can easily be determined that the last element of the nth row is n^2+n-1.

Now, the sum of all the elements of the nth row

= (n^2 - n+ 1) + (n^2 - n+ 3)+......+ (n^2 + n - 1)

In the above series, the first term(f) = n^2-n=1; last term (l) = n^2+n-1; common difference(d) = 2; and so:

The number of terms(t) in the series

= [(n^2 + n -1)- (n^2 - n+ 1)]/2 + 1

= n

Consequently, the required sum

= (t/2)*(f+s)

= (n/2)*(2*n^2)

= n^3

*Edited on ***October 26, 2007, 5:45 am**