An Integer Product (Posted on 2022-05-26)

	Submitted by Brian Smith
Rating: 5.0000 (1 votes)
Solution:	(Hide)
P(n) = (4-2/1)*(4-2/2)*...*(4-2/n) Expand each term (4-2/k) into 2*(2k-1)/k P(n) = (2*1/1)*(2*3/2)*...*(2*(2n-1)/n) Simplify a bit P(n) = 2^n * 1*3*...*(2n-1) / n! Multiple the numerator and denominator by n! P(n) = n! * 2^n * 1*3*...*(2n-1) / (n! * n!) Combine n! and 2^n into a n-term product P(n) = 2*4*...*(2n) * 1*3*...*(2n-1) / (n! * n!) Then the numerator simplifies into (2n)! P(n) = (2n)!/(n! * n!) Now its easy to see that P(n) is just 2n choose n which is always an integer for natural number n.

	Subject	Author	Date
	No Subject	xdog	2022-05-27 09:06:32
	re(3): demonstration	Charlie	2022-05-27 06:49:06
	re(2): demonstration	Kenny M	2022-05-26 20:20:59
	re(2): demonstration	Charlie	2022-05-26 17:28:51
	re: demonstration	Kenny M	2022-05-26 16:37:42
	demonstration	xdog	2022-05-26 12:38:06
	verification as integers	Charlie	2022-05-26 10:43:32