Answer
See the detailed answer below.
Work Step by Step
First, we need to recall Equation 21.18, which is
\begin{cases}
&\lambda_m=\dfrac{4L}{m} \\ \\
& f_m=\dfrac{mv}{4L} \tag {$m=1,3,5,..$}
\end{cases}
We know that the standing wave in an open-closed tube must have a node at the closed side and an antinode at the open side.
And as we see in the figure below, the standing wavelength decreases by a factor of 2 in each next frequency.
In the cases below, we can see that,
$\lambda_1=4L$, $\lambda_2=\frac{4}{3}L$, $\lambda_3=\frac{4}{5}L$, $\lambda_4=\frac{4}{7}L$, and so on...
Hence,
$$\boxed{\lambda_m=\dfrac{4L}{m}}\tag{where $m=1,3,5,...$}$$
And by the same approach, we can see that the frequency is also given by
$$v=f_m\lambda_m$$
$$f_m=\dfrac{v}{\lambda_m}$$
Plugging from the boxed formula above,
$$\boxed{f_m=\dfrac{mv}{4L}}\tag{where $m=1,3,5,...$}$$
And these two boxed formulas are as same as Equation 21.18.
