Rearange an expression I am learning induction. At one step I have to show that:
$$ 1 - \frac{1}{(1+n)} + \frac{1}{(n+1)(n+2)} $$
can be transformed to
$$ 1 - \frac{1}{n+2} $$
Theese are the steps for the transformation, but I cant understand them:
$$ 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} = 1 - \frac{(n+2) -1}{(n+1)(n+2)} = 1 - \frac{n+1}{(n+1)(n+2)} = 1 - \frac{1}{n+2} $$
Could you explain to me what happens there? 
 A: $$\begin{align}1-\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}&=1-\frac{n+2}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)}\\&=1+\frac{-(n+2)}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)}\\&=1+\frac{-(n+2)+1}{(n+1)(n+2)}\\&=1+\frac{-(n+1)}{(n+1)(n+2)}\\&=1+\frac{-1}{n+2}\\&=1-\frac{1}{n+2}\end{align}$$
A: The first step is as follows:
$$1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} = 1 - \frac{(n+2) -1}{(n+1)(n+2)}$$
Consider the fractional part of the first expression; this is:
$$- \frac{1}{n+1} + \frac{1}{(n+1)(n+2)}$$
Notice that there is a common factor of $1/(n+1)$, so we will factor this out and obtain:
$$\frac{1}{(n+1)}\left(-1+\frac{1}{n+2}\right)$$
Write $-1=-\frac{n+2}{n+2}$. Then the expression inside the brackets becomes $\frac{(n+2) -1}{n+2}$. Multiplying this by $1/(n+1)$ gives $\frac{(n+2) -1}{(n+1)(n+2)}$.
Now $(n+2)-1=n+1$; but there's $(n+1)$ on both the numerator and the denominator, and therefore this fraction reduces to $1/(n+2)$. Putting it all together, this gives the string of equivalent expressions:
$$1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} = 1 - \frac{(n+2) -1}{(n+1)(n+2)} = 1 - \frac{n+1}{(n+1)(n+2)} = 1 - \frac{1}{n+2}$$
