Question

Question #13c21

Answer 1

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 `(n+1)` QED.

Examples:

`n=4 => (8!)/(4!)^2 = 40320/(24^2) = 70 = 5 * 14`
`n=6 => (12!)/(6!)^2 = 579001600/(720^2) = 924 = 7 * 132`