Summation of finite series: Let $f(r)$ be what? Find the sum of the first $n$ terms of
$\displaystyle \frac{1}{1\times4\times7}+\frac{1}{4\times7\times10}+\frac{1}{7\times10\times13}+...$
My working:
Let $\begin{align}\displaystyle\frac{1}{(3x-2)(3x+1)(3x+4)} &\equiv \frac{A}{3x-2}+\frac{B}{3x+1}+\frac{C}{3x+4}\\1&\equiv A(3x+1)(3x+4)+B(3x-2)(3x+4)+C(3x-2)(3x+1)\end{align}$
Let $\displaystyle x=\frac{-1}{3}, B=\frac{-1}{9}$
Let $\displaystyle x=\frac{-4}{3}, A=\frac{1}{18}$
Let $\displaystyle x=\frac{2}{3}, C=\frac{1}{18}$
Thus, $\displaystyle\frac{1}{(3x-2)(3x+1)(3x+4)}\equiv \frac{1}{18(3x-2)}-\frac{1}{9(3x+1)} +\frac{1}{18(3x+4)}$
So, I thought of using the method $$\sum_{r=1}^n u_r= \sum_{r=1}^n [f(r+1)-f(r)]=f(n+1)-f(1)$$ but I’m not sure if finding such a function is possible in this case.
Carrying on using the above method,
$$\sum_{r=1}^n \frac{1}{(3r-2)(3r+1)(3r+4)}$$
$$=\sum_{r=1}^n \Bigl(\frac{1}{18(3r-2)} -\frac{1}{9(3r+1)} + \frac{1}{18(3r+4)}\Bigr)$$
$$=\sum_{r=1}^n \Bigl(\frac{1}{18(3r-2)} -\frac{1}{18(3r+1)}\Bigr) -\sum_{r=1}^n\Bigl(\frac{1}{18(3r+1)} -\frac{1}{18(3r+4)}\Bigr)$$
But now I don’t know what to let $f(r)=$?
How to proceed?
 A: In your final line, the $r$ should be replaced by $x$ to get
$$\sum_{x=1}^n \Bigl(\frac{1}{18(3x-2)} - \frac{1}{18(3x+1)}\Bigr) -\sum_{x=1}^n\Bigl(\frac{1}{18(3x+1)} - \frac{1}{18(3x+4)}\Bigr) \tag{1}\label{eq1A}$$
In each summation, the term being subtracted (i.e., $\frac{1}{18(3x+1)}$ and $\frac{1}{18(3x+4)}$) is the same as the first term being added for the next summation value, so they cancel, showing each is a Telescoping series. Thus, all of the terms cancel except for the first term minus the last term, giving
$$\begin{equation}\begin{aligned}
& \left(\frac{1}{18(3(1) - 2)} - \frac{1}{18(3n + 1)}\right) - \left(\frac{1}{18(3(1) + 1)} - \frac{1}{18(3n + 4)}\right) \\
& = \frac{1}{18} - \frac{1}{72} + \frac{1}{54n + 72} - \frac{1}{54n + 18} \\
& = \frac{1}{24} - \frac{54}{(54n + 72)(54n + 18)} \\
& = \frac{1}{24} - \frac{1}{6(3n + 4)(3n + 1)} \\
& = \frac{(3n + 4)(3n + 1)}{24(3n + 4)(3n + 1)} - \frac{4}{24(3n + 4)(3n + 1)} \\
& = \frac{9n^2 + 3n + 12n + 4 - 4}{24(3n + 4)(3n + 1)} \\
& = \frac{9n^2 + 15n}{24(3n + 4)(3n + 1)} \\
& = \frac{3n^2 + 5n}{8(3n + 4)(3n + 1)}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
A: Do telescopic summation after doing the partial fractions:
$$T_k=\frac{1}{(3k-2)(3k+1)(3k+4)}$$ $$\implies  18 T_k=\left(\frac{1}{3k-2}-\frac{1}{3k+1}\right)-\left( \frac{1}{3k+1}-\frac{1}{3k+4)}\right)$$
Let $F_k=\frac{1}{3k-2}$
$$S_n=\sum_{k=1}^{n} T_k= \frac{1}{18}(\sum_{k=1}^n [F_k-F_{k+1}]-\sum_{k=1}^n [F(k+1)-F(k+2)]).$$
$$\implies S_n=\frac{1}{18}([F_1-F_{n+1}]-[F_2-F_{n+2}])$$
You may complete it now.
