Generalize the equality $\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{n\cdot(n+1)}=\frac{n}{n+1}$ I'm reading a book The Art and Craft of Problem Solving. I've tried to conjecture a more general formula for sums where denominators have products of three terms. I've "got my hands dirty", but don't see any regularity in numerators.
Please, write down your ideas.

This kind of technique calls partial fractions and can be applied to several similar problems.
If you want a very powerfull techinque, try to read a book called A=B.
Using Mathematica (not only three terms, but using k ones)
$\displaystyle\sum_{x=1}^{n}\prod_{i=0}^{k}\frac{1}{x+i}=\frac{-k \Gamma (k+2) \Gamma (n+1)-n \Gamma (k+2) \Gamma (n+1)-\Gamma (k+2)
   \Gamma (n+1)+k \Gamma (k+n+2)+\Gamma (k+n+2)}{k \Gamma (k+2) \Gamma
   (k+n+2)}$
$\Gamma(n+1)=n!$
