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

Home > Numbers
Infinite Product (Posted on 2013-06-24) Difficulty: 2 of 5
Find the value of the infinite product
P=7/9*26/28*63/65*....*(k3-1)/(k3+1)*...

No Solution Yet Submitted by Danish Ahmed Khan    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Analytic solution Comment 3 of 3 |
consider the kth product

(k - 1) / (k + 1)

factor the numerator:

(k-1)(k+k+1)


consider the k+1th product:

((k+1) - 1) / ((k+1) + 1)

expand the denominator:
 
k + 3k + 3k + 2

and factor: (k+2)(k + k + 1)  

So the quadratic term in the kth numerator cancels the quadratic term in the k+1th demoninator, leaving a product (k-1)*k*(k+1)... in the numerator, and a product (k+2)(k+3)... in the denominator, AND the very first denominator which has no prior numerator to cancel.

All of the (k+n) terms in the denominator cancel corresponding terms in the numerator, leaving just three terms in the numerator without matches: (k-1)*k*(k+1), and the very first denominator is still left. That means the product overall is equal to:

(K - K) / (K + 1) where K is the very first value of k: 2 for this problem.

Substituting, the product is (8-2) / (9) = 6/9 = 2/3


  Posted by Paul on 2013-06-24 18:07:53
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 (6)
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