7・10 A fishing boat is fishing in the territorial sea of a foreign country without permission. Each time it casts a net, it causes the same value of loss to the country's fishing yield. The probability of the boat being detained by the foreign coast patrol during each net casting is equal to $1 / k$, where $k$ is a natural number of that country. Assume that the events of the boat being detained or not during each net casting are independent of the previous fishing process. If the fishing boat is detained by the foreign coast patrol, all the fish caught will be confiscated, and it will not be allowed to fish again in the future. The captain plans to leave the foreign territorial sea after casting the net for the $n$th time. Because the possibility of the fishing boat being detained by the foreign coast patrol cannot be ruled out, the fishing income is a random variable. Find the number $n$ that maximizes the expected value of the fishing income.
[Solution] The probability that the fishing boat is not detained during the first net casting is $1-\frac{1}{k}$. Since the events of being detained or not during each net casting are independent, the probability of not being detained after casting the net $n$ times is $\left(1-\frac{1}{k}\right)^{n}$. Therefore, the expected value of the profit from casting the net $n$ times is
$$f(n)=w n\left(1-\frac{1}{k}\right)^{n}$$

Here $w$ is the profit from casting the net once.
The problem is to determine the natural number $n$ that maximizes $f(n)$.
From the expression (1) of $f(n)$, we know
$$\begin{aligned}
f(n+1) & =w(n+1)\left(1-\frac{1}{k}\right)^{n+1} \\
& =w n\left(1-\frac{1}{k}\right)^{n} \cdot\left(1-\frac{1}{k}\right)\left(\frac{n+1}{n}\right) \\
& =f(n)\left(1-\frac{1}{k}\right)\left(1+\frac{1}{n}\right) \\
& =f(n)\left[1+\frac{(k-1)-n}{k n}\right] .
\end{aligned}$$

Since $1+\frac{(k-1)-n}{k n} \geqslant 1$ is equivalent to $k-1-n \geqslant 0$ or $n \leqslant k-1$, therefore
$$\begin{array}{l}
f(n+1)>f(n), \text { when } n=1,2, \cdots, k-2 \text {; } \\
f(n+1)=f(n), \text { when } n=k-1 \text {; } \\
f(n+1)<f(n), \text { when } n=k, k+1, \cdots \text {; }
\end{array}$$

Therefore, $f$ reaches its maximum value when $n=k-1$.