29. Let $m$ and $n$ be two positive integers that satisfy
$$
\frac{m}{n}=\frac{1}{10 \times 12}+\frac{1}{12 \times 14}+\frac{1}{14 \times 16}+\cdots+\frac{1}{2012 \times 2014}
$$

Find the smallest possible value of $m+n$.
29. Answer: 10571
$$
\begin{aligned}
\frac{m}{n}=\frac{1}{4} \sum_{k=5}^{1006} \frac{1}{k(k+1)} & =\frac{1}{4} \sum_{k=5}^{1006} \frac{1}{k}-\frac{1}{k+1} \\
& =\frac{1}{4}\left(\frac{1}{5}-\frac{1}{1007}\right) \\
& =\frac{501}{10070} .
\end{aligned}
$$

Since $\operatorname{gcd}(501,10070)=1$, we have $m+n=10571$.