Hello Erfan,
We are given to evaluate:
$$S_n=\sum_{r=1}^n\left(\frac{r}{(2r-1)(2r+1)(2r+3)} \right)$$
We are instructed to use partial fraction decomposition on the summand, and so we assume it may be expressed in the form:
$$\frac{r}{(2r-1)(2r+1)(2r+3)}=\frac{A}{2r-1}+\frac{B}{2r+1}+\frac{C}{2r+3}$$
Using the Heaviside cover-up method, we find:
$$A=\frac{\frac{1}{2}}{(2)(4)}=\frac{1}{16}$$
$$B=\frac{-\frac{1}{2}}{(-2)(2)}=\frac{1}{8}$$
$$C=\frac{-\frac{3}{2}}{(-4)(-2)}=-\frac{3}{16}$$
Hence:
$$\frac{r}{(2r-1)(2r+1)(2r+3)}=\frac{1}{16(2r-1)}+\frac{1}{8(2r+1)}-\frac{3}{16(2r+3)}$$
Factoring out $$\frac{1}{16}$$, we may write the sum as follows:
$$S_n=\frac{1}{16}\left(\sum_{r=1}^n\left(\frac{1}{2r-1} \right)+2\sum_{r=1}^n\left(\frac{1}{2r+1} \right)-3\sum_{r=1}^n\left(\frac{1}{2r+3} \right) \right)$$
Re-indexing the sums, we have:
$$S_n=\frac{1}{16}\left(\sum_{r=1}^n\left(\frac{1}{2r-1} \right)+2\sum_{r=2}^{n+1}\left(\frac{1}{2r-1} \right)-3\sum_{r=3}^{n+2}\left(\frac{1}{2r-1} \right) \right)$$
Pulling off terms so that the indices of summation are the same for all three sums, we find:
$$S_n=\frac{1}{16}\left(\left(1+\frac{1}{3}+ \sum_{r=3}^n\left(\frac{1}{2r-1} \right) \right)+\right.$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left.\left(\frac{2}{3}+2 \sum_{r=3}^n\left(\frac{1}{2r-1} \right)+\frac{2}{2n+1} \right)+\left(-3 \sum_{r=3}^n\left(\frac{1}{2r-1} \right)-\frac{3}{2n+1}-\frac{3}{2n+3} \right) \right)$$
Now the sums all add to zero, and we are left with:
$$S_n=\frac{1}{16}\left(2-\frac{1}{2n+1}-\frac{3}{2n+3} \right)=\frac{n(n+1)}{2(2n+1)(2n+3)}$$
