How to find $\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)}$? I tried to solve it, but the answer I got was different from the answer given.


Answer given:

$$\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)} = \frac{n(2n+1)}{4(2n-1)(2n+3)}$$


My working:

$$\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)}$$
$$= \sum_{r=1}^{n} \Biggl[\frac{1}{16(2r-1)}+\frac{1}{8(2r+1)}-\frac{3}{16(2r+3)}\Biggl]$$
$$= \frac{1}{16} + \frac{1}{24} - \require{cancel} \cancel{\frac{3}{80}}$$
$$+ \frac{1}{48} + \require{cancel} \cancel{\frac{1}{40}} - \require{cancel} \cancel{\frac{3}{112}}$$
$$+ \require{cancel} \cancel{\frac{1}{80}} + \require{cancel} \cancel{\frac{1}{56}} - \require{cancel} \cancel{\frac{3}{144}}$$
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$$+ \require{cancel} \cancel{\frac{1}{16(2n-3)}} + \require{cancel} \cancel{\frac{1}{8(2n-1)}} - \frac{3}{16(2n+1)}$$
$$+ \require{cancel} \cancel{\frac{1}{16(2n-1)}} + \frac{1}{8(2n+1)} - \frac{3}{16(2n+3)}$$
$$= \frac{1}{16} + \frac{1}{24} + \frac{1}{48} - \frac{3}{16(2n+1)} + \frac{1}{8(2n+1)} - \frac{3}{16(2n+3)}$$
$$= \frac{2n^{2}+6n+3}{4(2n+1)(2n+3)}$$

 A: Substituting $n=1$, the "correct answer" gives a wrong result as mentioned in the comments.
Indeed  $$\sum_{r=1}^{1}\frac{r}{(2r-1)(2r+1)(2r+3)}=\frac{1}{1\times3\times5}=\frac{1}{15}$$ but $$\frac{n(2n+1)}{4(2n-1)(2n+3)}\big|_{n=1}=\frac{3}{4\times1\times5}=\frac{3}{20}\neq \frac{1}{15}.$$
Also you should have:
$$\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)}$$
$$=\frac{1}{16}+\frac{1}{24}+\frac{1}{48}-\frac{3}{16(2n+1)}+\frac{1}{8(2n+1)}-\frac{3}{16(2n+3)}$$
$$=\frac{1}{8}-\frac{3(2n+3)}{16(2n+1)(2n+3)}+\frac{2(2n+3)}{16(2n+1)(2n+3)}-\frac{3(2n+1)}{16(2n+1)(2n+3)}$$
$$=\frac{2(2n+1)(2n+3)}{16(2n+1)(2n+3)}-\frac{2(4n+3)}{16(2n+1)(2n+3)}=\frac{(2n+1)(2n+3)-(4n+3)}{8(2n+1)(2n+3)}$$
$$=\frac{4n^{2}+8n+3-4n-3}{8(2n+1)(2n+3)}=\frac{n(n+1)}{2(2n+1)(2n+3)}.$$
A: Using partial fraction decmposition
$$\frac{r}{(2r-1)(2r+1)(2r+3)}=\frac{1}{8 (2 r+1)}-\frac{3}{16 (2 r+3)}+\frac{1}{16 (2 r-1)}$$ that you can rewrite
$$\frac 18 \left(\frac{1}{2 r+1}-\frac{1}{2 r+3} \right)+\frac 1{16} \left(\frac{1}{2 r-1}-\frac{1}{2 r+3} \right)$$ There would be a lot of telescoping
A: $$
\begin{align}
&\sum_{r=1}^n\frac{r}{(2r-1)(2r+1)(2r+3)}\\
&=\sum_{r=1}^n\left(\frac1{16(2r-1)}+\frac1{8(2r+1)}-\frac3{16(2r+3)}\right)\tag1\\
&=\frac1{16}\sum_{r=1}^n\left(\frac1{2r-1}-\frac1{2r+1}\right)+\frac3{16}\sum_{r=1}^n\left(\frac1{2r+1}-\frac1{2r+3}\right)\tag2\\
&=\frac1{16}\left(1-\frac1{2n+1}\right)+\frac3{16}\left(\frac13-\frac1{2n+3}\right)\tag3\\[3pt]
&=\frac18-\frac18\frac{4n+3}{(2n+1)(2n+3)}\tag4
\end{align}
$$
Explanation:
$(1)$: partial fractions
$(2)$: group for telescoping
$(3)$: telescope the sums
$(4)$: collect terms
A: Hint:
Another way:
Let $$\dfrac r{(2r-1)(2r+1)(2r+3)}=f(r)-f(r+1)$$  where $f(m)=\dfrac{am+b}{(2m-1)(2m+1)}$
so that $$\sum_{r=1}^n\dfrac r{(2r-1)(2r+1)(2r+3)}=\cdots=f(n+1)-f(1)$$
Now comparing the numerator of $f(r)-f(r+1)$ with $r,$
$$2a=1, a+4b=0\iff b=-\dfrac a4$$
