Try this beautiful problem from** Sum of Series from SMO**, Singapore Mathematics Olympiad, 2013.

## Sum of Series from SMO, 2013

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 .

- 10570
- 10571
- 16001
- 20000

**Key Concepts**

Greatest Common Divisor (gcd)

Sequence and Series

Number Theory

## Check the Answer

But try the problem first…

Answer: 10571

Singapore Mathematics Olympiad

Challenges and Thrills – Pre – College Mathematics

## Try with Hints

First Hint …………………………

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 ………………

Second Hint

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…………………

Final Step

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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