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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

But try the problem first…

Source

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.