Find the sum to n terms of the series $\frac{1}{1.2.3}+\frac{3}{2.3.4}+\frac{5}{3.4.5}+\frac{7}{4.5.6}+\cdots$.. Question : 
Find the sum to n terms of the series $\frac{1}{1.2.3}+\frac{3}{2.3.4}+\frac{5}{3.4.5}+\frac{7}{4.5.6}+\cdots$ 
What I have done : 
nth term of numerator and denominator is $2r-1$ and $r(r+1)(r+2)$ respectively.
Therefore the nth term of given series is : 
$\frac{2r-1}{r(r+1)(r+2)} =\frac{A}{r}+\frac{B}{r+1}+\frac{C}{r+2}$ .....(1)
By using partial fraction : 
and solving for A,B and C we get A = 1/2, B = -1, C =1/2
Putting the values of A,B and C in (1) we get : 
$\frac{1}{2r}-\frac{1}{r+1}+\frac{1}{2(r+2)}$
But by putting $r =1,2,3, \cdots$  I am not getting the answer. Please guide how to solve this problem . Thanks.

HINT:
If the $r$th term
$$T_r=\frac1{2r}-\frac1{r+1}+\frac1{2(r+2)}$$
$$2T_r=\frac1r-\frac2{r+1}+\frac1{(r+2)}=\left(\underbrace{\frac1r-\frac1{r+1}}\right)-\left(\underbrace{\frac1{r+1}-\frac1{r+2}}\right)$$
Recognize the two Telescoping series
