All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers
An Odd Pyramid (Posted on 2003-10-14) Difficulty: 3 of 5
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.

See The Solution Submitted by DJ    
Rating: 4.1667 (12 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Alternative Methodology Comment 21 of 21 |
(In reply to Puzzle Solution by K Sengupta)

Let G(x) = Sum of all the numbers from the first row to the nth
row.
g(x) = sum of all the numbers in the xth row.
S(x) = sum to x terms of the series (1+3+5+..)

Now, we observe that the xth term of the series 1+3+5... is (2x-1)

Thus, S(x) = (x/2)(1+2x-1) = x^2
Now, G(n) = S(1+2+...+n) = S[(n^2+n)/2) = n^2*(n+1)^2/4,
and, G(n-1) = S(1+2+...+(n-1)) = S[(n^2-n)/2) = n^2*(n-1)^2/4.

Consequently,
g(n) = G(n) - G(n-1) = (n^2)/4*[(n+1)^2 - (n-1)^2] = (n^2)/4*(4n) = n^3

Therefore, the required sum of the numbers in the nth row is n^3.


  Posted by K Sengupta on 2007-11-22 04:57:21
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (0)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information