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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.

No Solution Yet Submitted by K Sengupta    
Rating: 2.0000 (1 votes)

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Solution Full solution Comment 2 of 2 |
The sequence x(n) goes 2, 6, 20, 70, 252, 924, 3432
The OIES gives the formula x(n) = (2n)! / (n!)^2 from which it is clear that every term in the denominator will reduce to 1, leaving an integer.

Proof of the formula:
Show it works for n=1.  (2*1)! / (1!)^2 = 2

Assume it works for x(n) and show it works for x(n+1)

x(n+1) = 2(2(n+1)-1)*x(n)/(n+1)
=[2(2n+1)*(2n)!] / [(n+1)*(n!)^2]
multiply by (n+1)/(n+1)
=[(2n+2)(2n+1)(2n)!] / [(n+1)^2*(n!)^2]
=(2n+2)! / ((n+1)!)^2



  Posted by Jer on 2010-10-04 15:01:41
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