Simplify the expression $(2n)!/(2n+2)!$

Question

I'm a little confused as to how $(2n)!/(2n+2)!$ looks when written out.

Basically I'm trying to visualise it so that I know how to cancel this and like terms in future.

Answer 1

Note that:

$$(2n+2)! = (2n+2) \cdot (2n + 1) \cdot \underbrace{2n \cdot (2n - 1) \cdot (2n - 2) \dots \cdot 2 \cdot 1}_{=(2n)!}$$

Which means $$(2n+2)! = (2n+2) \cdot (2n+1) \cdot (2n)!$$

So when dividing $(2n+2)!$ by $(2n)!$ only those first two factors of $(2n+2)!$ remain (in this case in the denominator).

Answer 2

Hint: One very useful property of factorials is that

$$(N + 1)! = (N + 1) N! \implies \frac{(N + 1)!}{N!} = (N + 1)$$

Similarly, using the fact that $(N + 2)! = (N + 2)(N + 1) N!$ will help simplify the desired quotient.

Answer 3

$\frac{(2n)(2n-1)(2n-2)...1}{(2n+2)(2n+1)(2n)(2n-1)(2n-2)...1} = \frac{1}{(2n+2)(2n+1)}$

Answer 4

$$ (2n+2)! = (2n+2)\cdot(2n+1)\cdot \underbrace{2n \cdot (2n - 1) \cdot (2n - 2) \cdot \space ... \space\cdot (2 \cdot 2) \cdot (2 \cdot 1}_{=(2n)!} $$ $$ \implies \frac {(2n+2)!}{(2n)!} (because\space the \space 2n!s \space are \space cancelling \space out)= (2n+2) \cdot (2n+1) $$ $$ \implies\frac {(2n)!}{(2n+2)!} = \frac {1} {\frac {(2n+2)!}{(2n)!}} \space [= \space ({\frac {(2n+2)!}{(2n)!}})^{-1}] \space because \space we \space know \space that \space \frac {p}{q} = \frac {1}{\frac {q}{p}} $$ $$ \implies \frac {(2n)!}{(2n+2)!} = \frac {1}{(2n+1)(2n+2)} = \frac {1}{4n^2+6n+2)} $$ $$ \implies \frac {(2n)!}{(2n+2)!} = \frac {1}{4n^2+6n+2} \space [={(4n^2+6n+2)}^{-1}] $$

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So that everyone can understand it, I summarized it again and added the last step. :) Have fun! :D