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 reply to re: solution by Charlie)

DARNIT... yeah Charlie, you're right. It's a common one though... and so I figured I could gloss over it...

So, in the interest of completeness, here you are:

Like many other similar sequences... we can use the calculus of differences.
The sequence we are trying to generate is:
1, 3, 7, 13, 21, ...
If we show the DIFFERENCES (of each number to the next) in the next row we have:
2, 4, 6, 8, ...
Do it once more... and we have:
2, 2, 2, ....

Aha.... it is in the second row that we found a constant number. Well, we all know (don't we?), that this means that the sequence can be generated by a second order equation. (An² + Bn + C)

so, let's solve for A, B, and C.
Let's use n=1, 2, 3 to set up the equations:
(1) A + B + C = 1
(2) 4A + 2B + C = 3
(3) 9A + 3B + C = 7

Subtract (1) from (2) and we have:
3A + B = 2 let's call this (4)

Subtract (2) from (3) and we have:
5A + B = 4 let's call this (5)

subtract (4) from (5) and we have:
2A = 2
Therefore, A = 1
If A = 1, then B = -1
If A = 1 and B = -1, then C = 1

So.... putting these constants back into the original generating function, we have:

1n² + -1n + 1

And this is what we wanted to show.

Posted by SilverKnight on 2003-10-14 14:35:35