Finding the sum of a series, don't understand how they get from one step to the next 
Can some one please explain how they get from the red boxed step to the blue boxed one?
 A: First of all, it should be obvious that they factored the $3$ out of the whole thing.
Secondly, we will note that the remainder is given as:
$$(\frac12-\frac14)+(\frac13-\frac15)+(\frac14-\frac16)+\dots(\frac1{k}-\frac1{k+2})$$
Rewriting it without parenthesis,
$$\frac12-\frac14+\frac13-\frac15+\frac14-\frac16+\frac15-\frac17+\dots\frac1k-\frac1{k+2}$$
Now if you look closely, you will note what I have highlighted.
$$\frac12\color{red}{-\frac14}+\frac13\color{blue}{-\frac15}\color{red}{+\frac14}\color{green}{-\frac16}\color{blue}{+\frac15}\color{orange}{-\frac17}+\dots\frac1k-\frac1{k+2}$$
Each term that I highlighted corresponds to another term of the same color.  If you notice, these two terms will cancel each other.
After all of the cancellations occur, the only remaining terms are
$$\frac11+\frac13-\frac1{k+1}-\frac1{k+2}$$
This type of summation is called a telescoping series.
A: $$\frac12-\frac14$$
$$\frac13-\frac15$$
$$\frac14-\frac16$$
$$\frac15-\frac17$$
$$・$$
$$・$$
$$・$$
$$\frac1{k-3}-\frac1{k-1}$$
$$\frac1{k-2}-\frac1{k}$$
$$\frac1{k-1}-\frac1{k+1}$$
$$\frac1k-\frac1{k+2}$$
$\frac14, \frac15, \frac1{k-1}, \frac1k$ are vanished.hope for help.
