How to figure out $\frac{((k+1)!)^{2}}{(2(k+1))!} \times \frac{(2k)!}{((k)!)^{2}} = \frac{(k+1)^{2}}{(2k+2)(2k+1)}$

Question

$$\frac{((k+1)!)^{2}}{(2(k+1))!} \times \frac{(2k)!}{((k)!)^{2}} = \frac{(k+1)^{2}}{(2k+2)(2k+1)}$$

What is the method for figuring out this product?

What I can discern:

1. The factorial was taken out of the first numerator.

2. The second numerator was cancelled.

What are some of the rules for simplifying this? Thank you for your help.

Answer 1

$\frac{((k+1)!)^{2}}{(2(k+1))!} \times \frac{(2k)!}{((k)!)^{2}}$

$\frac{(2k)!}{(2(k+1))!} \times \frac{((k+1)!)^{2}}{((k)!)^{2}}$

$\frac{(2k)!}{(2k)!(2k+1)(2k+2)} \times \frac{(k!(k+1))^{2}}{((k)!)^{2}}$

$\frac{1}{(2k+1)(2k+2)} \times \frac{(k+1)^{2}}{1}$

$\frac{(k+1)^{2}}{(2k+1)(2k+2)}$

Answer 2

Another way.

$\begin{array}\\ \frac{((k+1)!)^{2}}{(2(k+1))!} \times \frac{(2k)!}{((k)!)^{2}} &=\frac{(k!(k+1))^{2}}{(2k+2)!} \times \frac{(2k)!}{((k)!)^{2}}\\ &=\frac{(k!)^2(k+1)^{2}}{(2k)!(2k+1)(2k+2)} \times \frac{(2k)!}{((k)!)^{2}}\\ &=\frac{(k+1)^{2}}{(2k+1)(2k+2)} \times \frac{1}{1}\\ &=\frac{(k+1)^{2}}{(2k+1)(2(k+1))} \qquad\text{(this can be simplified a little more)}\\ &=\frac{k+1}{2(2k+1)} \\ \end{array} $