Find Sum of n terms of the Series: $ \frac{1}{1\times2\times3} + \frac{3}{2\times3\times4} + \frac{5}{3\times4\times5} + \cdots $ I want to find the sum to n terms of the following series.
$$ \frac{1}{1\times2\times3} + \frac{3}{2\times3\times4} + \frac{5}{3\times4\times5} + \frac{7}{4\times5\times6} + \cdots $$
I have found that the general term of the sum is
\begin{equation}
\sum\limits_{i=1}^n \frac{2i - 1}{i(i+1)(i+2)}
\end{equation}
so the nth term must be
\begin{equation}
\frac{2n - 1}{n(n+1)(n+2)}
\end{equation}
Now.. how should i evaluate the sum of the n terms? What is the method i must use in order to do so? Thank you in advance.
 A: $$\displaystyle \frac{2n-1}{n(n+1)(n+2)} = \frac{2}{(n+1)(n+2)}-\frac{1}{n(n+1)(n+2)} = A-B$$
Now $$\displaystyle A = \frac{2}{(n+1)(n+2)} = 2\left[\frac{(n+2)-(n+1)}{(n+1)(n+2)}\right] = 2\left[\frac{1}{n+1}-\frac{1}{n+2}\right]$$
And $$\displaystyle B = \frac{1}{n(n+1)(n+2)} = \frac{1}{2}\left[\frac{(n+2)-(n)}{n(n+1)(n+2)}\right] = \frac{1}{2}\left[\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right]$$
$$\displaystyle = \frac{1}{2}\left\{\left[\frac{1}{n}-\frac{1}{n+1}\right]-\left[\frac{1}{n+1}-\frac{1}{n+2}\right]\right\}$$
Now $$\displaystyle \sum_{r=1}^{n}S_{n} = 2\sum_{r=1}^{n}\left[\frac{1}{r+1}-\frac{1}{r+2}\right]-\frac{1}{2}\sum_{r=1}^{n}\left[\frac{1}{r}-\frac{1}{r+1}\right]+\frac{1}{2}\sum_{r=1}^{n}\left[\frac{1}{r+1}-\frac{1}{r+2}\right]$$
Now Use Telescopic Sum
A: $$\begin{align}
\sum\limits_{i=1}^n \frac{2i - 1}{i(i+1)(i+2)}
&=\sum_{i=1}^n\frac {Ai+B}{i(i+1)}-\frac{A(i+1)+B}{(i+1)(i+2)}\\
&=\sum_{i=1}^n\frac{2i-\frac12}{i(i+1)}-\frac{2(i+1)-\frac12}{(i+1)(i+2)}\\
&=\frac34-\frac{4n+3}{2(n+1)(n+2)}\qquad\text{by telescoping}\\
&=\frac{n(3n+1)}{4(n+1)(n+2)}\qquad\blacksquare
\end{align}$$
