Hard Telescoping Series
Write $$\dfrac1{(2n+1)(2n+3)(2n+5)} =\\ \dfrac1{2n+3} \left(\frac{1}{(2n+1)(2n+5)}\right) =\\ \dfrac1{4(2n+3)} \left(\frac{1}{2n+1}-\frac{1}{2n+5}\right) =\\ \dfrac14 \left(\frac{1}{(2n+1)(2n+3)}-\frac{1}{(2n+3)(2n+5)}\right) =a_n-a_{n+1},\\ \text{for } \ a_n=\dfrac14 \left(\frac{1}{(2n+1)(2n+3)}\right) $$ to make it easy (telescopic with two factors).
$$\dfrac1{(2n+1)(2n+3)(2n+5)} = \dfrac18 \dfrac1{2n+1} - \dfrac14 \dfrac1{2n+3} + \dfrac18 \dfrac1{2n+5}$$ The above is obtained using partial fractions. The idea behind partial fractions in this case is to write $\dfrac1{(2n+1)(2n+3)(2n+5)}$ as $$\dfrac{A}{(2n+1)} + \dfrac{B}{(2n+3)} + \dfrac{C}{(2n+5)}$$ The goal is to find $A$, $B$ and $C$. Since $$\dfrac{A}{(2n+1)} + \dfrac{B}{(2n+3)} + \dfrac{C}{(2n+5)} = \dfrac1{(2n+1)(2n+3)(2n+5)}$$ we get that $$\dfrac{A(2n+3)(2n+5) + B(2n+1)(2n+5) + C(2n+1)(2n+3)}{(2n+1)(2n+3)(2n+5)} = \dfrac1{(2n+1)(2n+3)(2n+5)}$$ Hence, we need $A$, $B$ and $C$ such that $$A(2n+3)(2n+5) + B(2n+1)(2n+5) + C(2n+1)(2n+3) = 1$$ for all $n$. Since this is true for all $n$, plug in $n = -\dfrac12$ to get that $$8A = 1 \implies A = \dfrac18$$ Now, plug in $n = -\dfrac32$ to get that $$-4B = 1 \implies B = -\dfrac14$$ and finally, plug in $n = -\dfrac52$ to get that $$8C = 1 \implies C = \dfrac18$$ Hence, we get that $$\dfrac1{(2n+1)(2n+3)(2n+5)} = \dfrac18 \dfrac1{2n+1} - \dfrac14 \dfrac1{2n+3} + \dfrac18 \dfrac1{2n+5}$$
Hence, \begin{align} \sum_{n=1}^{N} \dfrac1{(2n+1)(2n+3)(2n+5)}& = \dfrac18 \cdot \dfrac1{3} - \dfrac14 \cdot \dfrac15 + \dfrac18 \cdot \dfrac17\\ & + \dfrac18 \cdot \dfrac1{5} - \dfrac14 \cdot \dfrac17 + \dfrac18 \cdot \dfrac19\\ & + \dfrac18 \cdot \dfrac1{7} - \dfrac14 \cdot \dfrac19 + \dfrac18 \cdot \dfrac1{11}\\ & + \dfrac18 \cdot \dfrac1{9} - \dfrac14 \cdot \dfrac1{11} + \dfrac18 \cdot \dfrac1{13}\\ & + \cdots \end{align} \begin{align} \sum_{n=1}^{N} \dfrac1{(2n+1)(2n+3)(2n+5)} & = \dfrac18 \cdot \dfrac13 + \left( \dfrac18 - \dfrac14\right)\cdot \dfrac15 + \left( \dfrac18 - \dfrac14 + \dfrac18 \right) \cdot \dfrac17\\ & + \left( \dfrac18 - \dfrac14 + \dfrac18 \right) \cdot \dfrac19 + \left( \dfrac18 - \dfrac14 + \dfrac18 \right) \cdot \dfrac1{11}\\ & + \left( \dfrac18 - \dfrac14 + \dfrac18 \right) \cdot \dfrac1{2n+1} + \left( \dfrac18 - \dfrac14 \right) \cdot \dfrac1{2n+3} + \dfrac18 \cdot \dfrac1{2n+5}\\ & = \dfrac18 \cdot \dfrac13 + \left( \dfrac18 - \dfrac14\right)\cdot \dfrac15 + \left( \dfrac18 - \dfrac14 \right) \cdot \dfrac1{2n+3} + \dfrac18 \cdot \dfrac1{2n+5}\\ & = \dfrac1{24} - \dfrac1{40} - \dfrac18 \cdot \dfrac1{2n+3} + \dfrac18 \cdot \dfrac1{2n+5}\\ & = \dfrac1{60} - \dfrac14 \cdot \dfrac1{(2n+3)(2n+5)} \end{align}
First, note that the telescoping series method only works on certain fractions. In particular, in order for the fractions to cancel out, we need the numerators to be the same. The typical example of telescoping series (for partial fractions) is
$$ \frac {1}{n(n+1)} = \frac {1}{n} - \frac {1}{n+1} \Rightarrow \sum_{i=1}^n \frac {1}{i(i+1)} = \sum_{i=1}^n \frac {1}{i} - \frac {1}{i+1} = \frac {1}{1} - \frac {1}{n+1}$$
If the numerators do not cancel out completely, then telescoping series will no longer work. For example, since we have
$$ \frac {2}{n} - \frac {1}{n+1} = \frac {n+2}{n(n+1)},$$
We find that $\sum \frac {n+2}{n(n+1)}$ doesn't yield to the telescoping series method.
How do we extend this? The key idea is that the numerators must cancel out. For example, from $1-2+1=0$, we can create
$$ \frac {1}{n} - \frac {2}{n+1} + \frac {1}{n+2} = \frac {2}{n(n+1)(n+2)}$$
which then tells us that
$$\sum_{i=1}^n \frac {2}{i(i+1)(i+2)} = \sum_{i=1}^n\frac {1}{i} - \frac {2}{i+1} + \frac {1}{i+2} = \frac {1}{1} - \frac {1}{2} + \frac {1}{n+1} - \frac {1}{n+2}.$$
You can easily see how to create more telescoping series sums using this idea. For example, what can we do with $3-4+1 = 0$ and the fraction:
$$ \frac 3 n - \frac 4 {n+1} + \frac 1 {n+2} = \frac {2(n+3)}{n(n+1)(n+2)}?$$
Edit: So, it sounds to me that you do not really see how telescoping series works. Let me expand it out, so you could understand it better. Let's first take the typical example of $\sum \frac {1}{i(i+1)}$. We can see that $\frac {1}{i(i+1)} = \frac {1}{i} - \frac {1}{i+1}$. What does this tell us? We have
$$\begin{array}{llllllllll} \frac {1}{1 \times 2} & = & \frac {1}{1} & - \frac {1}{2} \\ \frac {1}{2 \times 3} & = & & +\frac {1}{2} & - \frac {1}{3} \\ \frac {1} {3 \times 4} & = & & & + \frac {1}{3} & - \frac {1}{4} \\ \vdots & = \vdots \\ \frac {1}{(n-1) \times n} = &&&&&&& +\frac {1}{n-1} &- \frac {1}{n}\\ \frac {1}{n \times (n+1)} = &&&&&&& +\frac {1}{n} &- \frac {1}{n+1}\\ \end{array}$$
Now, summing up the LHS, we get $\sum_{i=1}^n \frac {1}{i(i+1)}$ as intended. Let's sum up the RHS according to each column. Then, we clearly see that lots of things cancel out, leaving us with $\frac 1 1 - \frac 1 {n+1}$.
We know do the same with $\frac {1}{n} - \frac {2}{n+1} + \frac {1}{n+2} = \frac {2}{n(n+1)(n+2)}$. Let's write it out as:
$$\begin{array}{llllllllll} \frac {2}{1 \times 2 \times 3} & = & \frac {1}{1} & - \frac {2}{2} & + \frac {1}{3}\\ \frac {2}{2 \times 3 \times 4} & = & & +\frac {1}{2} & - \frac {2}{3} & + \frac {2}{4}\\ \frac {2} {3 \times 4 \times 5} & = & & & + \frac {1}{3} & - \frac {2}{4} & + \frac {1}{5} \\ \vdots & = \vdots \\ \frac {2}{n \times (n+1) \times (n+2)} = &&&&&&& +\frac {1}{n-1} &- \frac {2}{n} &+ \frac {1}{n+1}\\ \frac {2}{n \times (n+1) \times (n+2)} = &&&&&&&& +\frac {1}{n} &- \frac {2}{n+1} + \frac {1}{n+2}\\ \end{array}$$
Now, can you tell what we get when we sum up the RHS over the columns? You should see that we get $\frac {1}{1} - \frac {1}{2} - \frac {1}{n+1} + \frac {1}{n+2}$.