<s>[INST] 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
 [/INST] 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
$$\sum_{r=1}^n T_r=\frac{n^2+3n}{4(n+2)}+\frac1{n+2}-\frac12=\frac{n^2+3n+4-2(n+2)}{4(n+2)}=\frac{n^2+n}{4(n+2)}$$
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