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

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

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  • $\begingroup$ @postmortes sorry, now it's ok $\endgroup$ – ami_ba Sep 3 '18 at 6:31
  • $\begingroup$ Since you have only $n$ terms you can neaten the sum up into $$\sum_{k=0}^{n-1} \frac{k+1}{(n-k)(n-k+1)(n-k+2)}$$ $\endgroup$ – postmortes Sep 3 '18 at 6:33
  • $\begingroup$ @ami_ba: You should try the decomposition into partial fractions. This is the standard way. $\endgroup$ – Marian G. Sep 3 '18 at 6:38
  • $\begingroup$ Hint: $T_r = \frac{n+3}{r(r+1)(r+2)} - \frac{r+2}{r(r+1)(r+2)}$. The first term has a constant (not dependent on $r$) in the enumerator, so try to "telescope" $\sum_{r=1}^n\frac{1}{r(r+1)(r+2)}$ first. Then simplify the second term and try to telescope that as well. $\endgroup$ – Ingix Sep 3 '18 at 6:38
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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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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 $$\begin{split} S_n &= \frac 1 2 \sum_{r=1}^n(n-r+1)(u_r-u_{r+1})\\ &= \frac 1 2 \bigg (\sum_{r=1}^n(n-r+1)u_r-\sum_{r=2}^{n+1} (n-r+2)u_r\bigg)\\ &= \frac 1 2 \bigg ( nu_1-u_{n+1} -\sum_{r=2}^n u_r \bigg)\\ &= \frac 1 2 \bigg ( nu_1 -\sum_{r=2}^{n+1} u_r \bigg)\\ &= \frac 1 2 \bigg (nu_1 - \frac 1 2 + \frac 1 {n+2} \bigg) \\ &= \frac 1 2 \bigg (\frac {n-1}2 + \frac 1 {n+2} \bigg) \\ \end{split} $$

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