If $\sqrt{\frac{3}{1} \cdot \frac{5}{3} \cdot \frac{7}{5} \cdot \cdots \cdot \frac{2 n+1}{2 n-1}}=9$, then the value of $n$ is
(A) 38
(B) 1
(C) 40
(D) 4
(E) 39
Since

$$
\sqrt{\frac{3}{1} \cdot \frac{5}{3} \cdot \frac{7}{5} \cdot \cdots \cdot \frac{2 n+1}{2 n-1}}=9
$$

then

$$
\frac{3}{1} \cdot \frac{5}{3} \cdot \frac{7}{5} \cdot \cdots \cdot \frac{2 n+1}{2 n-1}=81
$$

In the product on the left-hand side, the numerator in each fraction but the last appears as the denominator in the next fraction and so can be divided out. (This is called a "telescoping product".) After doing this cancellation, all that is left is the denominator of 1 from the first fraction and the numerator of $2 n+1$ from the last fraction. Therefore, $\frac{2 n+1}{1}=2 n+1=81$ or $n=40$.

ANSWER: (C)