Let-the-sum-n-1-9-1-n-n-1-n-2-written-in-its-lowest-terms-be-p-q-Find-the-value-of-q-p-

Question Number 20523 by Tinkutara last updated on 27/Aug/17

Let the sum \underset{n = 1}{\sum^{9}} \frac{1}{n (n + 1) (n + 2)} written

in its lowest terms be \frac{p}{q} . Find the value

of q - p .

Answered by ajfour last updated on 27/Aug/17

S = \frac{1}{2} \underset{n = 1}{\sum^{9}} \frac{(n + 2) - n}{n (n + 1) (n + 2)}

= \frac{1}{2} \underset{n = 1}{\sum^{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}) + \dots .

\dots . + (\frac{1}{9.10} - \frac{1}{10.11})]

= \frac{1}{2} [\frac{1}{2} - \frac{1}{110}] = \frac{27}{110}

\Rightarrow \frac{p}{q} = \frac{27}{110}

q - p = 83 .

Commented by Tinkutara last updated on 28/Aug/17

Thank you very much Sir!