12. For each positive integer $n$, we call $f(n)$ the product of all odd natural numbers less than or equal to $2n+1$ (for example, $f(4)=1 \cdot 3 \cdot 5 \cdot 7 \cdot 9$). We then set $g(n)=\frac{n}{f(n)}$. What can be said about the sum $S=g(1)+g(2)+\cdots+g(30)$?
(A) $S \leq 0.35$
(B) $0.35 < S \leq 0.45$
(C) $0.45 < S \leq 0.51$
(D) $S > 0.51$

## Numerical answer problems - 5 points
12. The answer is $(\mathbf{C})$. Let $S(m)=g(1)+g(2)+\cdots+g(m)$. Notice that

$$
\begin{aligned}
g(m)=\frac{m}{1 \cdot 3 \cdots(2 m+1)} & =\frac{1}{2} \frac{2 m}{1 \cdot 3 \cdots(2 m+1)}=\frac{1}{2} \frac{2 m+1-1}{1 \cdot 3 \cdots(2 m+1)} \\
& =\frac{1}{2}\left(\frac{1}{1 \cdot 3 \cdots(2 m-1)}-\frac{1}{1 \cdot 3 \cdots(2 m+1)}\right)
\end{aligned}
$$

Calling $h(m)=\frac{1}{2} \frac{1}{1 \cdot 3 \cdots(2 m+1)}$, we have just obtained that $g(m)=h(m-1)-h(m)$. Therefore, $S(m)=g(1)+g(2)+g(3)+\cdots+g(m-1)+g(m)=h(0)-h(1)+h(1)-h(2)+h(2)-$ $h(3)+\cdots+h(m-2)-h(m-1)+h(m-1)-h(m)$, which means $S(m)=h(0)-h(m)$. In the particular case,

$$
S(30)=\frac{1}{2}\left(1-\frac{1}{1 \cdot 3 \cdots(2 \cdot 30+1)}\right) .
$$

Returning to the general case, $S(30)=S0$; moreover, $\frac{1}{2} \frac{1}{1 \cdot 3 \cdots(2 \cdot 30+1)}0.49$.