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.

I have the following recipe in mind, see how far it helps (leave pointers in this regards as comments):
Let $a_1, a_2, a_3, \cdots, a_n, \cdots$ be the terms of an $A.P$. Let $d$ be the common difference of the given $A.P$. We are interested to find the sum for some $r \in \mathbb{N}$. $$\sum_{k=1}^n \dfrac{1}{a_k a_{k+1} \cdots a_{k+r-1}}$$
Let us denote the sum of the series as $S_n$ and the n-th term of the series, $T_n$. Consider the following definition of the new entity that I'll call $V_n$. (This is not at all a mystery: $V_n$ is obtained by dropping the first of the $r$ entries in the denominator of $T_n$)
$$V_n:=\dfrac{1}{a_{n+1} \cdots a_{n+r-2} a_{n+r-1}}$$
Therefore, $$V_{n-1}:=\dfrac{1}{a_n  \cdots a_{n+r-3} a_{n+r-2}}$$
Now (I'll leave the computation that goes here!), you'll have $$V_n-V_{n-1}=T_n(a_n-a_{n+r-1})$$ $$T_n=\dfrac{1}{d(r-1)} \cdot (V_{n-1}-V_n)$$Substituting various values for $n$, you have equations for $T_1, T_2, \cdots, T_n$. Adding these, and noting that common difference is the difference between two consecutive terms taken in an (appropriate!) order, you have, $$S_n=\dfrac{1}{(r-1)(a_2-a_1)}(\dfrac{1}{a_1\cdots a_{r-1}}-\dfrac{1}{a_{n+1} \cdots a_{n+r-1}})$$
For the problem at hand, set $a_1=1,~d=1,~r=3 $. You'll get $$\sum_{k=1}^n{\dfrac{1}{k(k+1)(k+2)}}=\dfrac{1}{4}-\dfrac{1}{(n+1)(n+2)}$$
Have fun doing for $r=4, \cdots$. Hope this helps.
