Answer
See below.
Work Step by Step
Proofs using mathematical induction consists of two steps:
1) The base case: here we prove that the statement holds for the first natural number.
2) The inductive step: assume the statement is true for some arbitrary natural number $n$ larger than the first natural number, then we prove that then the statement also holds for $n + 1$.
Hence here:
1) For $n=1: \frac{1}{1(2)}=\frac{1}{1+1}$.
2) Assume for $n: \frac{1}{1(2)}+\frac{1}{2(3)}+...+\frac{1}{n(n+1)}=\frac{n}{n+1}$. Then for $n+1$:
$\frac{1}{1(2)}+\frac{1}{2(3)}+...+\frac{1}{n(n+1)}+\frac{1}{(n+1)(n+2)}=\frac{n}{(n+1)}+\frac{1}{(n+1)(n+2)}=\frac{n(n+2)}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)}=\frac{n^2+2n+1}{(n+1)(n+2)}=\frac{(n+1)^2}{(n+1)(n+2)}=\frac{(n+1)}{(n+2)}=\frac{(n+1)}{(n+1)+1}$
Thus we proved what we wanted to.