How is $\frac{\left(2\left(n+1\right)\right)!}{\left(n+1\right)!}\cdot \frac{n!}{\left(2n\right)!}$ simplified like that?

Question

I am trying to solve some ratio tests. I came across to my notes and i can't figure out how is that equation simplified. Well, how is $\frac{\left(2\left(n+1\right)\right)!}{\left(n+1\right)!}\cdot \frac{n!}{\left(2n\right)!}$ simplified to $\frac{\left(2n+1\right)\left(2n+2\right)}{n+1}$

Answer 1

\begin{align} \frac{\left(2\left(n+1\right)\right)!}{\left(n+1\right)!}\cdot \frac{n!}{\left(2n\right)!} &= \left[ \frac{\left(2\left(n+1\right)\right)!}{\left(2n\right)!} \right]\cdot \frac{n!}{\left(n+1\right)!} \\ &= \frac{(2n)! (2n+2)(2n+1)}{(2n)!} \frac{n!}{n! (n+1)}\\ &= \frac{(2n+1)(2n+2)}{n+1} \end{align}

Answer 2

Hint: Try to simplify this: $$\dfrac{(n+1)!}{n!}$$By using: $n!=1 \cdot 2 \cdot 3 \cdots n.$

Answer 3

Note that: $$(2(n+1))!=(2n+2)!$$ $$=(2n)!\times (2n+1)\times (2n+2)$$ giving us: $$\frac{(2(n+1))!}{(2n)!}\times \frac{n!} {(n+1)!} $$ $$= \frac{(2n)!\times (2n+1) \times (2n+2)}{(2n)!}\times \frac{n!} {n! \times (n+1)}$$ $$=\,? $$

Answer 4

$[1]:\frac{(2(n+1))!}{(2n)!}$=$\frac{(2n+2)!}{(2n)!}$=$\frac{(2n+2)(2n+1)(2n)!}{(2n)!}=(2n+2)(2n+1)=2(n+1)(2n+1)$
$[2]:\frac{(n)!}{(n+1)!}=\frac{(n)!}{(n+1)(n)!}=\frac{1}{n+1}$

Multiply 1 by 2:

$\frac{(2(n+1))!}{(2n)!}.\frac{(n)!}{(n+1)!}=\frac{(2n+2)(2n+1)}{n+1}=2(n+1)(2n+1)(\frac{1}{n+1})=4n+2$

Answer 5

$$\frac{\left(2\left(n+1\right)\right)!}{\left(n+1\right)!}\cdot \frac{n!}{\left(2n\right)!}=\frac{n!}{(n+1)!}\cdot\frac{(2n+2)!}{(2n)!}=\frac{n!}{(n+1)\cdot n!}\cdot\frac{(2n+2)\cdot (2n+1)\cdot(2n)!}{(2n)!}=\frac{(2n+2)\cdot (2n+1)}{n+1}$$