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?
 A: I'm sure this answer is not rigorous, but perhaps someone can make it so.
By manipulating factorials we get
$$\frac{1}{K}=\frac{1}{2^{100}}\binom{100}{50}=P(X\,{=}\,50)\ ,$$
where $X$ is a binomial random variable with $n=100$ and $p=\frac{1}{2}$.  Approximating $X$ by a normal random variable $Y$ in the usual way, we have $Y\sim N(50,5^2)$ and so
$$\frac{1}{K}\approx P(49.5\,{<}\,Y\,{<}\,50.5)=P(-0.1\,{<}\,Z\,{<}\,0.1)
  \approx0.0796\ .$$
Hence
$$K\approx12.56$$
and so $\lfloor K\rfloor=12$.
A: I could not get any ideas of getting the answer analytically.
So I just wrote the matlab program as below
format rat
Initial=1
i=1;
count=0;
while i<100,
Initial = Initial*(i+1)/i;
i=i+2;
count=count+1;
a(count+1)= Initial;
end;
Got the answer 12 after flooring the end result.
A: Note that this answer is not completely rigorous, but it was too fun to pass up.

$$K^2 = 101 \cdot \frac{2 \cdot 2 \cdot 4 \cdot 4 \cdot \dots \cdot 100 \cdot 100}{1 \cdot 3 \cdot 3 \cdot 5 \cdot \dots \cdot 99 \cdot 101}$$
Now, $$\frac{2 \cdot 2 \cdot 4 \cdot 4 \cdot \dots}{1 \cdot 3 \cdot 3 \cdot 5 \cdot \dots} = \pi / 2$$
(This is known as the Wallis product)
So $K$ is approximately $\sqrt{101 \pi / 2}$ and $\lfloor K \rfloor = 12$
A: Note that $K = \dfrac{2 \cdot 2 \cdot 4 \cdot 4 \cdots 100 \cdot 100}{1 \cdot 2 \cdot 3 \cdot 4 \cdots 99 \cdot 100} = \dfrac{2^{100}(50!)^2}{100!} = \dfrac{2^{100}}{\dbinom{100}{50}}$
It can be shown that the Central Binomial Coefficent satisfies: 
$\left(1-\dfrac{1}{8n}\right)\dfrac{2^{2n}}{\sqrt{\pi n}} \le \dbinom{2n}{n} \le \dfrac{2^{2n}}{\sqrt{\pi n}}\left(1-\dfrac{1}{9n}\right)$
for all $n \ge 1$. 
Thus, $12.56105 \approx \dfrac{450}{449}\sqrt{50\pi} \le \dfrac{2^{100}}{\dbinom{100}{50}} \le \dfrac{400}{399}\sqrt{50\pi} \approx 12.56456$
Therefore, $\lfloor K \rfloor = 12$.
A: 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 $n \gt 5$.
