Question #13c21
1 Answer
Oct 17, 2017
We have:
((2n)!)/(n!)^2 = ( 1.2.3...n(n+1)(n+2)...(2n) ) / ( (1.2.3...n)(1.2.3...n) )
\ \ \ \ \ \ \ \ \ = ((n+1)(n+2)...(2n) ) / ( (1.2.3...n) )
\ \ \ \ \ \ \ \ \ = (n+1) * ((n+2)...(2n) ) / ( (1.2.3...n) )
So, we have shown that the expression has a factor of
Examples:
n=4 => (8!)/(4!)^2 = 40320/(24^2) = 70 = 5 * 14
n=6 => (12!)/(6!)^2 = 579001600/(720^2) = 924 = 7 * 132
