<s>[INST] Sum of a special type of series Is there a general method to derive the formula (without using mathematical induction) for the sum of the series of type $1*2+2*3+3*4+4*5+...+n(n+1)$ or perhaps $\frac{1}{(2*3*4)}+\frac{1}{(3*4*5)}+\frac{1}{(5*6*7)}+...+\frac{1}{(n*(n+1)*(n+2))}$?
I am a high school student so an elaborate explanation with example would help me best. I am looking for the formulas of those two series but I am more interested in the method of derivation of the formulas and also any general method by which such type of problems can be solved.
 [/INST] Given that the rising and falling factorials are defined  as
$$
\begin{array}{l}
 n^{\,\overline {\,m\,} }  = n\left( {n + 1} \right)\, \cdots \;\left( {n + m - 1} \right) \\ 
 n^{\,\underline {\,m\,} }  = n\left( {n - 1} \right)\, \cdots \;\left( {n - \left( {m + 1} \right)} \right) \\ 
 n^{\,\underline {\, - \,m\,} }  = \frac{1}{{\left( {n + m} \right)^{\,\underline {\,m\,} } }} = \frac{1}{{\left( {n + 1} \right)^{\,\overline {\,m\,} } }} \\ 
 \end{array}
$$
where here we consider $n$ and $m$ integers,
and that the forward finite difference (integer $q$) is defined as:
$$
\begin{array}{l}
 \Delta _{\,n} \;f(n) = f(n + 1) - f(n)\quad \quad \Delta _{\,n} ^q \;f(n) = \Delta _{\,n} \;\left( {\Delta _{\,n} ^{q - 1} \;f(n)} \right) \\ 
 \Delta _{\,n} ^{ - 1} \;f(n) = \sum\limits_{k\, = \,a}^{n - 1} {f(k)}  + c \\ 
 \end{array}
$$
then for the falling and rise factorials we have in general
$$
\begin{array}{l}
 \Delta _{\,n} ^q \,n^{\,\overline {\,m\,} }  = m^{\,\underline {\,q\,} } \,\left( {n + q} \right)^{\,\overline {\,m - q\,} } \quad \quad \Delta _{\,n} ^q \,n^{\,\underline {\,m\,} }  = m^{\,\underline {\,q\,} } \,n^{\,\underline {\,m - q\,} }  \\ 
 \sum\limits_{k\, = \,1}^{n - 1} {k^{\,\overline {\,m\,} } } \quad \left| {\; - 1 \ne m} \right.\,\quad  = \frac{1}{{m + 1}}\,\left( {\left( {n - 1} \right)^{\,\overline {\,m + 1\,} }  - 0^{\,\overline {\,m + 1\,} } } \right) = \frac{1}{{m + 1}}\,\left( {n - 1} \right)^{\,\overline {\,m + 1\,} }  \\ 
 \sum\limits_{k\, = \,1}^{n - 1} {k^{\,\underline {\, - m\,} } } \quad \left| {\;1 \ne m} \right.\,\quad  = \sum\limits_{k\, = \,1}^{n - 1} {\frac{1}{{\left( {k + 1} \right)^{\,\overline {\,m\,} } }}}  = \frac{1}{{1 - m}}\,\left( {n^{\,\underline {\,1 - m\,} }  - 1^{\,\underline {\,1 - m\,} } } \right) =  \\ 
  = \frac{1}{{1 - m}}\,\left( {\frac{1}{{\left( {n + 1} \right)^{\,\overline {\,m - 1\,} } }} - \frac{1}{{2^{\,\overline {\,m - 1\,} } }}} \right) \\ 
 \end{array}
$$
In the two particular cases you indicate, then
$$
\begin{array}{l}
 \sum\limits_{k\, = \,1}^n {k^{\,\overline {\,2\,} } }  = \sum\limits_{k\, = \,1}^n {k\left( {k + 1} \right)}  = \frac{1}{3}\,n^{\,\overline {\,3\,} }  \\ 
 \sum\limits_{k\, = \,1}^{n - 1} {k^{\,\underline {\, - 3\,} } }  = \sum\limits_{k\, = \,2}^n {\frac{1}{{k\left( {k + 1} \right)\left( {k + 2} \right)}}}  =  \\ 
  = \frac{1}{{1 - 3}}\,\left( {\frac{1}{{\left( {n + 1} \right)^{\,\overline {\,2\,} } }} - \frac{1}{{2^{\,\overline {\,2} } }}} \right) = \frac{1}{2}\,\left( {\frac{1}{6} - \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}} \right) \\ 
 \end{array}
$$
Besides the above, you may also consider the following.
The first type, can be translated into
$$
n\left( {n + 1} \right) = n^{\,\overline {\,2\,} }  = \sum\limits_{0\, \le \,k\, \le \,2} {\left[ \begin{array}{c}
 2 \\ 
 k \\ 
 \end{array} \right]n^{\,k} } 
$$
where $n^{\,\overline {\,2\,} }$ is the rising factorial and ${\left[ \begin{array}{c} 2\\ k\\  \end{array} \right]x^{\,k} } $
is the unsigned Stirling N. of 1st kind.
So the whole boils down into a combination of the sum of $n$ and $n^2$,
 (and for exponents greater than $2$ will involve Bernoulli numbers).
Concerning the second, instead, it can be splitted by partial fractions into
$$
\frac{1}{{n\left( {n + 1} \right)\left( {n + 2} \right)}} = \frac{1}{2}\frac{1}{n} - \frac{1}{{n + 1}} + \frac{1}{2}\frac{1}{{n + 2}}
$$
and thus solved in terms of harmonic numbers in general, while in this simple case, the sum will turn out in cancelling many of terms as pointed out by Mariuslp in his comment.
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