## 22. We Need Boys

The ruler of a certain country, for purely military reasons, wanted there to be more boys than girls among his subjects. Therefore, he decreed that there should not be more than one girl in any family. As a result, in this country, among the children of each woman, the last and only the last was a girl, because no woman, having given birth to a girl, dared to have more children. What proportion of boys made up the total number of children in this country?
22. Let $N$ be the number of mothers who no longer have children. How many of them have girls? One each, i.e., $N$. And how many have boys? Half of them have no boys at all, since the probability that the first child born is a girl is $1 / 2$. A quarter of the mothers have one boy, because $1 / 4 = 1 / 2 \cdot 1 / 2$ is the probability that the first child is a boy and the second is a girl. An eighth of the mothers have two boys (for similar reasons). A sixteenth of the mothers have three boys, and so on. In total, we get

\[
\begin{aligned}
& \frac{N}{4} \cdot 1 + \frac{N}{8} \cdot 2 + \frac{N}{16} \cdot 3 + \ldots + \frac{N}{2^{n+1}} \cdot n + \ldots = \\
& \quad = \frac{N}{4} \left[ \frac{1}{2^{0}} + \frac{2}{2^{1}} + \frac{3}{2^{2}} + \ldots + \frac{n}{2^{n-1}} + \ldots \right]
\end{aligned}
\]

110
boys. But it is known that

\[
1 + 2x + 3x^{2} + \ldots + nx^{n-1} + \ldots = \frac{1}{(1-x)^{2}}
\]

because this series is the derivative of the sum of a geometric progression

\[
1 + x + x^{2} + \ldots + x^{n} + \ldots = \frac{1}{1-x}
\]

By setting $x = 1 / 2$ in (*), we find

\[
\frac{1}{2^{0}} + \frac{2}{2^{1}} + \frac{3}{2^{2}} + \ldots + \frac{n}{2^{n-1}} = \ldots = \frac{1}{\left(1 - \frac{1}{2}\right)^{2}} = 4
\]

From which it follows that the number of boys is

\[
N / 4 \cdot 4 = N
\]

i.e., equal to the number of girls.

Thus, the proportion of boys in the total number of children is $1 / 2$. Despite the royal decree, it has not increased at all! ?[^21]