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sum of series $\displaystyle \frac{n}{1\cdot 2 \cdot 3}+\frac{n-1}{2\cdot 3\cdot 4}+\frac{n-2}{3\cdot 4 \cdot 5}+\cdots \cdots n$ terms

assuming $\displaystyle S_{n} =\frac{n}{1\cdot 2 \cdot 3}+\frac{n-1}{2\cdot 3\cdot 4}+\frac{n-2}{3\cdot 4 \cdot 5}+\cdots +\frac{n-(n-1)}{n\cdot n+1 \cdot n+2}$

$\displaystyle S_{n} = \sum^{n-1}_{r=0}\frac{n-r}{(r+1)\cdot (r+2) \cdot (r+3)}$

wan, t be able to solve after that , could some help me with this

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Hint:

$\dfrac{n-r}{(r+1)(r+2)(r+3)}=\dfrac{n+1-(r+1)}{(r+1)(r+2)(r+3)}$ $=(n+1)\cdot\dfrac1{(r+1)(r+2)(r+3)}-\dfrac1{(r+2)(r+3)}$

Now $\dfrac1{(r+2)(r+3)}=\dfrac{r+3-(r+2)}{(r+2)(r+3)}=?$

See Telescoping series

Again, $\dfrac2{(r+1)(r+2)(r+3)}=\dfrac{r+3-(r+1)}{(r+1)(r+2)(r+3)}=\dfrac1{(r+2)(r+1)}-\dfrac1{(r+2)(r+3)}$

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