If $K = \frac{2}{1}\times \frac{4}{3}\times \cdots \times \frac{100}{99}.$ Then value of $\lfloor K \rfloor$ 
Let $K = \frac{2}{1}\times \frac{4}{3}\times \frac{6}{5}\times \frac{8}{7}\times \cdots \times \frac{100}{99}.$ Then what is the value of $\lfloor K \rfloor$, where $\lfloor x \rfloor$ is the floor function?

My Attempt:
By factoring out powers of $2$, we can write
$$
\begin{align}
K &= 2^{50}\times \left(\frac{1}
{1}\times \frac{2}{3}\times \frac{3}{5}\times \frac{4}{7}\times \frac{5}{9}\times\cdots\times \frac{49}{97}\times \frac{50}{99}\right)\\
&= 2^{50}\cdot 2^{25}\times \left(\frac{1\cdot 3 \cdot 5\cdots49}{1\cdot 3 \cdot 5\cdots 49}\right)\times \left(\frac{1}{51\cdot 53\cdot 55\cdots99}\right)\\
&= \frac{2^{75}}{51\cdot 53\cdot 55\cdots99}
\end{align}
$$
How can I solve for $K$ from here?

If we make the problem more general and write $$\displaystyle K_n = \frac{2}{1}\times \frac{4}{3}\times \frac{6}{5}\times \frac{8}{7}\times \cdots \times \frac{2n}{2n-1}$$ the numerator is $2^n \Gamma (n+1)$ and the denominator is $\frac{2^n \Gamma \left(n+\frac{1}{2}\right)}{\sqrt{\pi }}$. So, $$K_n=\frac{\sqrt{\pi } \Gamma (n+1)}{\Gamma \left(n+\frac{1}{2}\right)}$$
Considering this expression for large values of $n$, we then have $$K_n=\sqrt{\pi } \sqrt{n}+\frac{1}{8} \sqrt{\pi } \sqrt{\frac{1}{n}}+\frac{1}{128}
   \sqrt{\pi } \left(\frac{1}{n}\right)^{3/2}-\frac{5 \sqrt{\pi }
   \left(\frac{1}{n}\right)^{5/2}}{1024}+O\left(\left(\frac{1}{n}\right)^3\right)$$ which implies $$\lfloor K_n \rfloor =\lfloor \sqrt{\pi n} \rfloor$$ which is verified for any value of $n \gt 5$.
