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.

  • $\begingroup$ $\phantom{}$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

Hint: $$ \frac1{n(n+1)(n+2)}=\frac12\left[\frac1{n(n+1)}-\frac1{(n+1)(n+2)}\right] $$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.