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.
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}