How do you simplify $\frac{(2 n + 2)!}{(2 n)!}$ $((2n+2)!)/((2n)!)$ ?

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Answer 1 · 2015-11-24T00:48:03

I found: $(2 n + 2) (2 n + 1)$ $(2n+2)(2n+1)$

Remembering that we can write:
$n! = n \cdot (n - 1)!$ $n! =n*(n-1)!$

we have (applying twice the above property to the numerator):
$((2n+2)!)/((2n)!) =((2n+2)(2n+1)cancel((2n)!))/(cancel((2n)!))=(2n+2)(2n+1)$

Check with $n=3$
$((2n+2)!)/((2n)!) =(8!)/(6!)=40320/720=56$
and:
$(2n+2)(2n+1)=8*7=56$

How do you simplify ((2n+2)!)/((2n)!)(2n+2)!(2n)!((2n+2)!)/((2n)!)?