Answer
See explanations.
Work Step by Step
Step 1. Prove the formula is true when $n=1$: $LHS=\frac{1}{1\times2}=\frac{1}{2}$, $RHS=\frac{1}{1+1}=\frac{1}{2}$, thus it is true for $n=1$.
Step 2. Assume the formula is true when $n=k$: we have $\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+... +\frac{1}{k(k+1)}=\frac{k}{k+1}$
Step 3. Prove it is true for $n=k+1$: $LHS=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+... +\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}=\frac{k(k+2)+1}{(k+1)(k+2)}=\frac{(k+1)^2}{(k+1)(k+2)}=\frac{k+1}{k+2}$
Use $n=k+1$ on the right side of the formula, we have $RHS=\frac{k+1}{k+2}$
Thus, we have $LHS=RHS$ which proves the formula is also true for $n=k+1$
Step 4. Conclusion: with mathematical induction, we have proved that the formula is true for all natural numbers n.
