calculus - Sum to $n$ terms the series $frac{1}{3cdot9cdot11}+frac{1}{5cdot11cdot13}+frac{1}{7cdot13cdot15}+cdots$.

Q:Sum to n terms the series :

$$\frac{1}{3\cdot9\cdot11}+\frac{1}{5\cdot11\cdot13}+\frac{1}{7\cdot13\cdot15}+\cdots$$

This was asked under the heading using method of difference and ans given was

$$S_n=\frac{1}{140}-\frac{1}{48}\left(\frac{1}{2n+3}+\frac{1}{2n+5}+\frac{1}{2n+7}-\frac{3}{2n+9} \right)$$

My Approach:First i get

$$U_n=\frac{1}{(2n+1)(2n+7)(2n+9)}$$
In order to make $U_n$ is the reciprocal of the product of factors in A.P i rewrite it
$$U_n=\frac{(2n+3)(2n+5)}{(2n+1)(2n+3)(2n+5)(2n+7)(2n+9)}=\frac{(2n+7)(2n+9)-48}{(2n+1)(2n+3)(2n+5)(2n+7)(2n+9)}=\frac{1}{(2n+1)(2n+3)(2n+5)}-\frac{48}{(2n+1)(2n+3)(2n+5)(2n+7)(2n+9)}$$
Then i tried to make $U_n=V_n-V_{n-1}$ in order to get $S_n=V_n-V_0$.But i really don't know how can i figure out this.Any hints or solution will be appreciated.
Thanks in advance.

Answer

Let

$$S_n=\sum_{k=1}^n\frac{1}{(2k+1)(2k+7)(2k+9)}.$$
By the partial fraction decomposition:
$$\frac{1}{(2k+1)(2k+7)(2k+9)}=\frac{1/48}{2k+1} -\frac{1/12}{2k+7}+\frac{1/16}{2k+9}$$
Then, after letting $O_n=\sum_{k=1}^n\frac{1}{2k+1}$, we have that
$$\begin{align} S_n&=\frac{O_n}{48}-\frac{O_{n+3}-O_3}{12}+\frac{O_{n+4}-O_4}{16}\\ &=\frac{4O_3-3O_4}{48}+\frac{O_n-4O_{n+3}+3O_{n+4}}{48}\\ &=\frac{1}{140}-\frac{1}{48}\left(O_{n+3}-O_n-3(O_{n+4}-O_{n+3})\right)\\ &=\frac{1}{140}-\frac{1}{48}\left(\frac{1}{2n+3}+\frac{1}{2n+5}+\frac{1}{2n+7}-\frac{3}{2n+9} \right). \end{align}$$