It's Only Slightly Bigger Than 1

Let $\mathcal{P}$ denote that product. We multiply numerator and denominator by $2 \times 4 \times 6 \times \ldots \times 100$ , which gives

$\begin{aligned} \mathcal{P} & = & \frac { \left ( 2 \times 4 \times 6 \times \ldots \times 100 \right )^2 }{ 100! } \\ & = & 2^{100} \times \frac { 50!^2 }{100!} \\ & = & 2^{100} \times \frac {1}{ {100 \choose 50} } \\ \end{aligned}$

Apply the asymptotic properties for Central Binomial Coefficient :

${2n \choose n } \sim \frac {4^n}{\sqrt{\pi n}}$

In this case, $n=50$

$\mathcal{P} \approx 2^{100} \times \frac { \sqrt{\pi \times 50} }{ 4^{50}} = 5\sqrt{2\pi}$

Hence the answer is $\left \lfloor 5\sqrt{2\pi} \right \rfloor = \boxed{12}$

Using the formulas for $\int\limits_{0}^{\pi/2}\sin^nxdx$ , for $n=99,100$ , and $101$ , we find that $\frac{98!!}{99!!}>\frac{99!!}{100!!}\frac{\pi}{2}>\frac{100!!}{101!!}$ . We rearrange things to see that $\frac{100\pi}{2}<(\frac{100!!}{99!!})^2<\frac{101\pi}{2}$ , so $12.5<\frac{100!!}{99!!}<12.6$ . The answer is $12$ .

Let $A = \frac{2}{1} \times \frac{4}{3} \times \frac{ 6}{5} \times \ldots \times \frac{ 98}{97} \times \frac{ 100}{99}.$ We will show that $144 < A^2 < 169$ .

For the lower bound, it suffices to show that

$\frac{ 2^2}{1 \times 3} \times \frac{ 4^2}{3 \times 5} \times \frac{ 6^2}{5 \times 7 } \times \ldots \times \frac{ 100^2}{99 \times 101} > \frac{ 144}{101}.$

This is true because the product of the first 3 terms is $\frac{ 256}{175} > \frac{ 144}{101}$ , and the rest of the terms are all greater than 1.

For the upper bound, it suffices to show that

$\frac{ 2 \times 4} { 3^2} \times \frac{ 4 \times 6}{5^2} \times \frac{ 6 \times 8 } { 7^2 } \times \ldots \times \frac { 98 \times 100} { 99^2} < \frac{ 169}{200} .$

Similarly, this is true because the product of the first 3 terms is $\frac{ 1024}{1225} < \frac{ 169}{200}$ , and the rest of the terms are all lesser than 1.

{product(2n), n=1 to 50}/{product(2n-1), n=1 to 50}

12.5645

WolframAlpha

Using MAPLE, I find that:

$\frac{2}{1} \times \frac{4}{3} \times \cdots \times \frac{100}{99} = \frac{158456325028528675187087900672}{12661418068195524166851562157}$

which is EXACT, and NOT APPROXIMATE (I hate approximations...). Then, one can compute the floor of this "large" fraction to be $12$ , as required.

class MyClass {

public static void main(String[ ] args) {


double a=1;


for(int n=1;n<=50;n++)


{


    a=a*(2*n)/(2*n-1);


}


System.out.println(""+a);


}

}

OUTPUT: 12.5645...