Try this beautiful problem from the Pre-RMO, 2017 based on Series.

## Series Problem – PRMO 2017

Let the sum \(\displaystyle\sum_{n=1}^{9}\frac{1}{n(n+1)(n+2)}\) written in its lowest terms be \(\frac{p}{q}\), find the value of q-p.

- is 107
- is 83
- is 840
- cannot be determined from the given information

**Key Concepts**

Series

Integers

Rearrangement of terms in series

## Check the Answer

But try the problem first…

Answer: is 83.

PRMO, 2017, Question 6

Calculus Vol 1 and 2 by Apostle

## Try with Hints

First hint

here \(\displaystyle\sum_{n=1}^{9}\frac{n+2-n}{2n(n+1)(n+2)}\)

=\(\frac{1}{2}\displaystyle\sum_{n=1}^{9}(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)})\)

Second Hint

=\(\frac{1}{2}(\frac{1}{(1)(2)}-\frac{1}{(2)(3)}\)

+\(\frac{1}{(2)(3)}-\frac{1}{(3)(4)}+….+\frac{1}{(9)(10)}-\frac{1}{(10)(11)})\)

=\(\frac{1}{2}(\frac{1}{2}-\frac{1}{110})\)

Final Step

\(\Rightarrow \frac{27}{110}\)

\(\Rightarrow q-p=110-27\)

=83.

## Other useful links

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

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