Sum of Series from SMO - 2013 - Problem Number 29

Let m and n be two positive integers that satisfy

\(\frac {m}{n} = \frac {1}{10\times 12} + \frac {1}{12 \times 14} + \frac {1}{14 \times 16} + \cdot +\frac {1}{2012 \times 2014} \)

Find the smallest possible value of m+n .

We can start this kind some by using the concept of series and sequence .......

In this problem we can see that the series as

\(\frac {m}{n}\) =\(\frac {1}{10 \times 12}\) +\(\frac {1}{12 \times 14}\) +\( \cdot \cdot\)+

\(\frac {1}{2012 \times 2014} \)

So sum of this series is

\(\frac {m}{n} = \frac {1}{4} \displaystyle\sum _{k = 5}^{1006} \frac {1}{k(k+1)}\)

Now do the rest of the sum ..................

If you are really stuck after the first hint here is the rest of the sum...............

From the above hint we can continue this problem by breaking the formula more we will get :

= \(\frac {1}{4} \displaystyle\sum_{k=5}^{1006} \frac {1}{k} - \frac {1}{k+1}\)

Now replacing by the values:

\(\frac {1}{4} (\frac {1}{5} - \frac {1}{1007}) \)

Please try to do the rest.....................

This is the last hint as well as the final answer....

If we continue after the last hint...

\(\frac {m}{n} = \frac {501}{10070}\)

Since gcd(501,10070) = 1

we can conclude by the values of m= 501 and n = 10070

So the sum is m+n = 10571 (Answer).

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