Answer
See explanations.
Work Step by Step
Step 1. Prove the formula is true when $n=1$: $LHS=1\times3=3$, $RHS=\frac{1(2)(9)}{6}=3$, thus it is true for $n=1$.
Step 2. Assume the formula is true when $n=k$: we have $1\times3+2\times4+3\times5+...+k(k+2)=\frac{k(k+1)(2k+7)}{6}$
Step 3. Prove it is true for $n=k+1$:
$LHS=1\times3+2\times4+3\times5+...+k(k+2)+(k+1)(k+3)=\frac{k(k+1)(2k+7)}{6}+(k+1)(k+3)=\frac{k(k+1)(2k+7)+6(k+1)(k+3)}{6}=\frac{(k+1)(2k^2+7k+6k+18)}{6}=\frac{(k+1)(2k^2+13k+18)}{6}$
$RHS=\frac{(k+1)(k+2)(2(k+1)+7)+1]}{6}=\frac{(k+1)(k+2)(2k+9)}{6}=\frac{(k+1)(k+2)(2k+9)}{6}=\frac{(k+1)(2k^2+13k+18)}{6}$
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.
