<s>[INST] Sum the series $\frac{3}{1⋅2⋅4}+\frac{4}{2⋅3⋅5}+ \frac{5}{3⋅4⋅6}+...\text{(upto n terms)}$ $\frac{3}{1⋅2⋅4}+\frac{4}{2⋅3⋅5}+ \frac{5}{3⋅4⋅6}+...\text{(upto n terms)}$
The general term seems to be
$$T_r= \frac{r+2}{r(r+1)(r+3)}.$$
I see no way to telescope this because the factors of the denominator of the general term are not in arithmetic progression.
Do I have to use something else? Or am I missing some easy manipulation?
 [/INST] We have that
$$a_r=\frac{r + 2}{r (r + 1) (r + 3)} = \frac2{3 r}-\frac1{2 (r + 1)} - \frac1{6 (r + 3)}=$$$$= \frac1{2r}-\frac1{2 (r + 1)}+ \frac1{6r} - \frac1{6 (r + 3)}=$$
$$= \frac12\left(\frac1{r}-\frac1{r + 1}\right)+ \frac16\left(\frac1{r} - \frac1{r + 3}\right)$$
and therefore
$$\sum_1^n a_r=\frac12 \sum_1^n \left(\frac1{r}-\frac1{r + 1}\right)+\frac16\sum_1^n\left(\frac1{r} - \frac1{r + 3}\right)=$$
$$=\frac12\left(1-\frac1{n+1}\right)+\frac16\left(1+\frac12+\frac13-\frac1{n+1}-\frac1{n+2}-\frac1{n+3}\right)=$$
$$=\frac{29}{36}-\frac2{3(n+1)}-\frac1{6(n+2)}-\frac1{6(n+3)}=$$
$$=\frac{29 n^3+138 n^2+157 n}{36(n+1)(n+2(n+3)}$$
See also the related Sum the series $\frac{n}{1⋅2⋅3}+\frac{n-1}{2⋅3⋅4}+ \frac{n-2}{3⋅4⋅5}+...\text{(upto n terms)}$.
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