Sum to n terms of the series $\frac{1}{\mathrm{1.2.3}}+\frac{3}{\mathrm{2.3.4}}+\frac{5}{\mathrm{3.4.5}}+\frac{7}{\mathrm{4.5.6}}+\dots$ is

Solution:

 $\frac{1}{\mathrm{1.2.3}}+\frac{3}{\mathrm{2.3.4}}+\frac{5}{\mathrm{3.4.5}}+\frac{7}{\mathrm{4.5.6}}+\dots$

We have $t_r=\frac{2r-1}{r(r+1)(r+2)}$ and 
$\begin{array}{l} t_{r+1}=\frac{2r+1}{(r+1)(r+2)(r+3)}\\ \Rightarrow r{t}_r=\frac{2r-1}{(r+1)(r+2)},(r+3)t_{r+1}=\frac{2r+1}{(r+1)(r+2)}\\ \Rightarrow r{t}_r-(r+3)t_{r+1}=\frac{-2}{(r+1)(r+2)}\end{array}$
$\begin{array}{l}\Rightarrow r{t}_r-(r+1)t_{r+1}+2\left(\frac{1}{r+1}-\frac{1}{r+2}\right)=2t_{r+1}\\ \Rightarrow \sum _{r=1}^{n-1} \left[r{t}_r-(r+1)t_{r+1}\right]+2\sum _{r=1}^{n-1} \left(\frac{1}{r+1}-\frac{1}{r+2}\right)=2\sum _{r=1}^{n-1} t_{r+1}\\ \Rightarrow t_1-n{t}_n+1-\frac{2}{n+1}=2\left(t_2+t_3+\dots +t_n\right)\\ \Rightarrow 3t_1-n{t}_n+1-\frac{2}{n+1}=2\sum _{r=1}^n t_r\\ \frac{1}{2}-\frac{n(2n-1)}{n(n+1)(n+2)}+\frac{n-1}{n+1}=2\sum _{r=1}^n t_r\end{array}$
$\begin{array}{l}\therefore 2\sum _{r=1}^n t_r=\frac{(n+1)(n+2)-2(2n-1)+2(n-1)(n+2)}{2(n+1)(n+2)}\\ \Rightarrow \sum _{r=1}^n t_r=\frac{n(3n+1)}{4(n+1)(n+2)}\end{array}$