NUMERO

We can rewrite this as $\frac{1}{1\times2} + \frac{1}{2\times3} + \frac{1}{3\times4} + ... + \frac{1}{(p-2)\times (p-1)}$. All of these terms are in the form of $\frac{1}{n(n+1)}$, so let's first try to simplify that.

$\frac{1}{n(n+1)}$

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

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

$\frac{1}{n} - \frac{1}{n+1}$

Now, we can rewrite all of out fractions as $(\frac{1}{1} - \frac{1}{2})+(\frac{1}{2} - \frac{1}{3})+(\frac{1}{3} - \frac{1}{4}) + ... +(\frac{1}{p-2} - \frac{1}{p-1})$

We can finally simplify this to $1 - \frac{1}{p-1}$ or $\mathbf{\frac{p-2}{p-1}}$, which is the final answer.

You could also solve this by induction, because you could notice that $\frac{1}{2}+\frac{1}{2\times3}=2/3$ and that  $\frac{1}{1\times2}+\frac{1}{2\times3} + \frac{1}{3\times4}=3/4$.