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

Home > Numbers > Sequences
Cross Product Equality Settlement (Posted on 2010-10-04) Difficulty: 2 of 5
A sequence {xn} is defined as:
x1 = 2, and:
n*xn = 2(2n-1)*xn-1, whenever n ≥ 2

Prove that xn is always an integer for every positive integer value of n.

See The Solution Submitted by K Sengupta    
Rating: 3.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Possible Solution | Comment 1 of 2

The easiest way to approach this is to look for n as a factor in the (n-1)th or earlier terms of x since the RHS is multiplied by x(n-1) and then divided by n, to produce xn. Then we see immediately that each successive 2(2n-1) incorporates a 'bow-wave' of numbers whose factors are potential divisors of the subsequent values of x, in ample time for them to be 'knocked out' again by being divided out by some n, so that there are always sufficient matching factors available for the division. For example, the factor 7 appears when n=4, well in advance of n = 7; and again at n = 11 (and every subsequent n = 7k-3) in good time to divide out the n's having 7 as a factor.

Accordingly, xn is always an integer.

Edited on October 4, 2010, 1:44 pm
  Posted by broll on 2010-10-04 13:35:44

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 (8)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information