Reducing fractions

Question

How can I get [if it's possible]: $$ 2- \frac{1}{n+1} $$ from this: $$2- \frac{1}{n} + \frac{1}{n(n+1)} $$

EDIT:

I started like this: $$2- \frac{1}{n} + \frac{1}{n(n+1)}= 2-\frac{n+1}{n(n+1)}+\frac{1}{n(n+1)} =2-\frac{n+2}{n(n+1)} $$

How do I proceed from here?

Answer 1

I started like this: $$2- \frac{1}{n} + \frac{1}{n(n+1)}= 2-\frac{n+1}{n(n+1)}+\frac{1}{n(n+1)} =2-\frac{n+2}{n(n+1)} $$

The mistake happened here:

$$2-\frac{n+1}{n(n+1)}+\frac{1}{n(n+1)} =2-\frac{n+2}{n(n+1)}$$

because you forgot the minus sign. You can rewrite the left hand side as

$$2 +\frac{-n-1}{n(n+1)} + \frac{1}{n(n+1)}$$ and continue from there.

Answer 2

There is no need to care about the term $2$, it suffices to establish

$$\frac1{n+1}=\frac1n-\frac1{n(n+1)}.$$

If you multiply by $n(n+1)$, this is

$$n=n+1-1.$$