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The question is like this: Let the sum $$\sum_{n=1}^9 \frac{1}{n(n+1)(n+2)}$$ written in its lowest terms be $\frac{p}{q}$. Find $p-q$ I tried to calculate it by putting in values 1 to 9 and actually calculating the value of the sum, but it was too long and I don't want to use a calculator. I messed with the expression for quite a while only to realize it was in vain. Please help. thanks in advance.

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  • $\begingroup$ $\phantom{}$TLSCPC $\endgroup$ – Jack D'Aurizio Dec 28 '17 at 17:43
  • $\begingroup$ @ Jack D'Aurizio: aramaic-english? $\endgroup$ – cgiovanardi Dec 29 '17 at 13:48
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Hint: $$ \frac1{n(n+1)(n+2)}=\frac12\left[\frac1{n(n+1)}-\frac1{(n+1)(n+2)}\right] $$

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  • $\begingroup$ Is the answer 83? $\endgroup$ – user167920 Dec 28 '17 at 17:06
  • $\begingroup$ Yes, that is correct up to a factor of $-1$. $\endgroup$ – robjohn Dec 28 '17 at 17:08
  • $\begingroup$ Oh yeah right it was numerator minus denominator. -83. My bad. $\endgroup$ – user167920 Dec 28 '17 at 17:17
  • $\begingroup$ I like this. I usually decompose it into 3 fractions. Never thought of doing this. $\endgroup$ – Elie Louis Dec 28 '17 at 17:36

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