1. If $\frac{2013 \times 2013}{2014 \times 2014+2012}=\frac{n}{m}$ (where $m$ and $n$ are coprime natural numbers), then the value of $m+n$ is ( ).
(A) 1243
(B) 1343
(C) 4025
(D) 4029
Answer: B.
In the exam, choosing the right method is crucial. For this problem, my first thought was to use induction. $\frac{2^{2}}{3^{2}+1}=\frac{2}{5} 、 \frac{3^{2}}{4^{2}+2}=\frac{1}{2} 、 \frac{4^{2}}{5^{2}+3}=\frac{4}{7} 、 \frac{5^{2}}{6^{2}+4}=\frac{5}{8}$, after writing the first three, I noticed that the second expression seemed out of place, so I wrote out the fourth one. Upon closer inspection, I realized that the second expression could be written as $\frac{3^{2}}{4^{2}+2}=\frac{3}{6}$. The pattern became clear: the numerator is a factor of the original numerator, and the denominator is 3 more than the numerator! The answer must be $\frac{2013}{2016}$, which is quite simple. The first question is usually an easy one involving the year, and the year 2013 is a familiar number, $2013=3 \times 11 \times 61$, which is a multiple of 3. Adding 3 still results in a multiple of 3, so the fraction can be simplified. Therefore, the final answer is $\frac{2013}{2016}=\frac{671}{672}$, so the correct choice is B!