Sum the series $\frac{n}{1⋅2⋅3}+\frac{n-1}{2⋅3⋅4}+ \frac{n-2}{3⋅4⋅5}+...\text{(upto n terms)}$ 
$\frac{n}{1⋅2⋅3}+\frac{n-1}{2⋅3⋅4}+ \frac{n-2}{3⋅4⋅5}+...\text{(upto n terms)}$

Clearly, we can see that $$T_r=\frac{n-r+1}{r(r+1)(r+2)}$$
Now, somehow, we have to make this telescoping.
But we do not have $n$ as a factor of the denominator, so we have to multiply the denominator by $n$ and after our manipulation, multiply the whole sum by $n$. But that too did not work.
PLease help
 A: We have that
$$T_r=\frac{n-r+1}{r(r+1)(r+2)}=\frac{n+1}{r(r+1)(r+2)}-\frac{r}{r(r+1)(r+2)}=$$$$=(n+1)\left(\frac{1}{2r}+\frac{1}{2(r+2)}-\frac{1}{r+1}\right)+\frac1{r+2}-\frac1{r+1}$$
and by telescoping we can see that the first term gives
$$(n+1)\left(\frac12+\frac14+\frac12\cdot 2\sum_{r=3}^n\left(\frac1r\right)+\frac12\frac1{n+1}+\frac12\frac1{n+2}-\sum_{r=2}^n\left(\frac1r\right)-\frac1{n+1}\right)$$
$$(n+1)\left(\frac{1}{4} -\frac12\frac1{n+1}+\frac12\frac1{n+2}\right)$$
$$(n+1)\left(\frac{(n+1)(n+2)-2(n+2)+2(n+1)}{4(n+1)(n+2)}\right)$$
$$\frac{n^2+3n}{4(n+2)}$$
and the second one
$$\frac1{n+2}-\frac12$$
therefore
$$\sum_{r=1}^n T_r=\frac{n^2+3n}{4(n+2)}+\frac1{n+2}-\frac12=\frac{n^2+3n+4-2(n+2)}{4(n+2)}=\frac{n^2+n}{4(n+2)}$$
A: Let $$u_r= \frac 1 r - \frac 1{r+1}$$
Then
$$\begin{split}
u_r-u_{r+1} &=  \frac 1 r - \frac 1{r+1} -\bigg( \frac 1 {r+1} - \frac 1{r+2}\bigg) \\
&= \frac 1 r - \frac 2{r+1} + \frac 1 {r+2} \\
&= \frac{ (r+2)(r+1) -2r(r+2) + r(r+1)}{r(r+1)(r+2)}\\
&= \frac 2 {r(r+1)(r+2)}\\
\end{split}$$
With that, the sum to compute is $$\begin{split}
S_n &= \frac 1 2 \sum_{r=1}^n(n-r+1)(u_r-u_{r+1})\\
&= \frac 1 2 \bigg (\sum_{r=1}^n(n-r+1)u_r-\sum_{r=2}^{n+1} (n-r+2)u_r\bigg)\\
&= \frac 1 2 \bigg ( nu_1-u_{n+1} -\sum_{r=2}^n u_r \bigg)\\
&= \frac 1 2 \bigg ( nu_1 -\sum_{r=2}^{n+1} u_r \bigg)\\
&= \frac 1 2 \bigg (nu_1 - \frac 1 2 + \frac 1 {n+2} \bigg) \\
&= \frac 1 2 \bigg (\frac {n-1}2 + \frac 1 {n+2} \bigg) \\
\end{split}
$$
