## Task B-2.3.

Calculate:

$$
\frac{\left(1+\frac{1}{3}\right)\left(1+\frac{1}{5}\right) \cdot \ldots \cdot\left(1+\frac{1}{2 n+1}\right) \cdot \ldots \cdot\left(1+\frac{1}{4047}\right)}{\sqrt{\left(1-\frac{1}{3^{2}}\right)\left(1-\frac{1}{5^{2}}\right) \cdot \ldots \cdot\left(1-\frac{1}{(2 n+1)^{2}}\right) \cdot \ldots \cdot\left(1-\frac{1}{4047^{2}}\right)}}
$$
## Solution.

Notice that the expression consists of the product of fractions of the form:

$$
\frac{\left(1+\frac{1}{2 n+1}\right)}{\sqrt{\left(1-\frac{1}{(2 n+1)^{2}}\right)}}
$$

Let's write it in a simpler form. Notice that the denominator is a difference of squares, and bring the numerator under the same root. Then we have:

$$
\begin{aligned}
& \frac{\left(1+\frac{1}{2 n+1}\right)}{\sqrt{\left(1-\frac{1}{(2 n+1)^{2}}\right)}}=\sqrt{\frac{\left(1+\frac{1}{2 n+1}\right)^{2}}{\left(1-\frac{1}{2 n+1}\right)\left(1+\frac{1}{2 n+1}\right)}}= \\
& =\sqrt{\frac{\left(1+\frac{1}{2 n+1}\right)}{\left(1-\frac{1}{2 n+1}\right)}}=\sqrt{\frac{2 n+2}{2 n}}=\sqrt{\frac{n+1}{n}} .
\end{aligned}
$$

Finally, if we represent each term of the original expression in this form, it follows that the desired product is equal to

$$
\sqrt{\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdot \ldots \cdot \frac{n+1}{n} \cdot \ldots \cdot \frac{2023}{2022} \cdot \frac{2024}{2023}}=\sqrt{2024}
$$