To solve the problem, we need to evaluate the given sum:

$\frac{m}{n} = \frac{1}{10 \times 12} + \frac{1}{12 \times 14} + \frac{1}{14 \times 16} + \dots + \frac{1}{2012 \times 2014}$

Each term in the sum can be expressed as a difference of fractions using the identity:

$\frac{1}{k(k+2)} = \frac{1}{2} \left(\frac{1}{k} - \frac{1}{k+2}\right)$

Applying this identity to each term:

$\frac{1}{10 \times 12} = \frac{1}{2} \left(\frac{1}{10} - \frac{1}{12}\right)$

$\frac{1}{12 \times 14} = \frac{1}{2} \left(\frac{1}{12} - \frac{1}{14}\right)$

And so on, until:

$\frac{1}{2012 \times 2014} = \frac{1}{2} \left(\frac{1}{2012} - \frac{1}{2014}\right)$

Therefore, the entire sum can be written as:

$\frac{m}{n} = \frac{1}{2} \left(\left(\frac{1}{10} - \frac{1}{12}\right) + \left(\frac{1}{12} - \frac{1}{14}\right) + \dots + \left(\frac{1}{2012} - \frac{1}{2014}\right)\right)$

Notice that this sum is a telescoping series, where most of the intermediate terms cancel out. Only the first term of the first fraction and the second term of the last fraction remain:

$\frac{m}{n} = \frac{1}{2} \left(\frac{1}{10} - \frac{1}{2014}\right)$

Simplifying this:

$\frac{m}{n} = \frac{1}{2} \times \frac{2014 - 10}{10 \times 2014} = \frac{2004}{2 \times 10 \times 2014} = \frac{1002}{10 \times 2014} = \frac{1002}{20140}$

Thus, we have:

$\frac{m}{n} = \frac{1002}{20140}$

To find the smallest possible value of $m + n$, we simplify the fraction $\frac{1002}{20140}$. We calculate the greatest common divisor (GCD) of 1002 and 20140:

• $1002 = 2 \times 3 \times 167$
• $20140 = 2 \times 10070 = 2 \times 2 \times 5035 = 2 \times 2 \times 5 \times 1007 = 2 \times 2 \times 5 \times 19 \times 53$

Since 1002 and 20140 have a common factor of 2, we divide both the numerator and denominator by 2:

$\frac{1002}{20140} = \frac{501}{10070}$

Now $\gcd(501, 10070) = 1$, so the fraction is in its simplest form. Therefore, $m = 501$ and $n = 10070$, and the smallest possible value of $m + n$ is:

$m + n = 501 + 10070 = 10571$

Thus, the correct answer is:

$\boxed{10571}$

Would you like further details or have any questions? Here are 5 related questions to expand your understanding:

1. How do you determine when to use telescoping series for simplification?
2. Can you explain why telescoping series work effectively in this context?
3. How would the problem change if the initial terms in the sum were different?
4. What other methods could be used to simplify sums involving products of consecutive integers?
5. How do you find the greatest common divisor (GCD) efficiently for larger numbers?

Tip: When dealing with fractions in a sequence, always check if the sum forms a telescoping series, as it greatly simplifies calculations.