<s>[INST] I need help with a Finite Series Problem:
Find the sum to $n$ terms of
\begin{eqnarray*}
\frac{1}{1\cdot 2\cdot 3} + \frac{3}{2\cdot 3\cdot 4} + \frac{5}{3\cdot 4\cdot 5} +
        \frac{7}{4\cdot 5\cdot 6}+\cdots \\
\end{eqnarray*}
Answer:
The way I see it, the problem is asking me to find this series:
\begin{eqnarray*}
S_n &=& \sum_{i=1}^{n} {a_i} \\
\text{with } a_i &=& \frac{2i-1}{i(i+1)(i+2)} \\
\end{eqnarray*}
We have:
\begin{eqnarray*}
S_n &=& S_{n-1}  + a_n \\
S_n &=& S_{n-1}  + \frac{2n-1}{n(n+1)(n+2)} \\
\end{eqnarray*}
I am tempted to apply the technique of partial fractions
but I believe there is no closed formula for a series of the of the form:
\begin{eqnarray*}
\sum_{i=1}^{n} \frac{1}{i+k} \\
\end{eqnarray*}
where $k$ is a fixed constant. Therefore I am stuck. I am hoping that somebody
can help me.
Thanks Bob
 [/INST] 
The good old way: compute a few terms and find a pattern.
$$\frac2{12},\frac7{24},\frac{15}{40},\frac{26}{60},\frac{40}{84}\cdots$$
By finite differences, we find that the numerators are $\dfrac{n(3n+1)}2$ and the denominators $2(n+1)(n+2)$.

Check:
$$S_1=\frac2{12}$$and
$$S_n-S_{n-1}=\frac{n(3n-1)}{4(n+1)(n+2)}-\frac{(n-1)(3n-4)}{4n(n+1)}=\frac{2n-1}{n(n+1)(n+2)}=a_n.$$
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