Series Problem | PRMO 2017 | Question 6

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.

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

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

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

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

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

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

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

=83.

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