Answer
See explanations.
Work Step by Step
Step 1. Prove the formula is true when $n=1$: $LHS=1\times2=2$, $RHS=\frac{1(2)(3)}{3}=2$, thus it is true for $n=1$.
Step 2. Assume the formula is true when $n=k$: we have $1\times2+2\times3+3\times4+...+k(k+1)=\frac{k(k+1)(k+2)}{3}$
Step 3. Prove it is true for $n=k+1$:
$LHS=1\times2+2\times3+3\times4+...+k(k+1)+(k+1)(k+2)=\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)=\frac{k(k+1)(k+2)+3(k+1)(k+2)}{3}=\frac{(k+1)(k+2)(k+3)}{3} $
Use $n=k+1$ for right side of the formula, we have $RHS=\frac{(k+1)(k+2)(k+3)}{3}$
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.
