Answer
See explanations.
Work Step by Step
Step 1. Recall the property of Fibonacci sequence, we have: $F_{n}=F_{n-1}+F_{n-2}$ and $F_1=F_2=1$
Step 2. Prove the statement is true when $n=1$: $F_{3}=F_2+F_1=2$ which is even, thus it is true for $n=1$.
Step 3. Assume the statement is true when $n=k$: we have $F_{3k}$ is even.
Step 4. Prove it is true for $n=k+1$: $F_{3(k+1)}=F_{3k+3}=F_{3k+2}+F_{3k+1}=F_{3k+1}+F_{3k}+F_{3k+1}=2F_{3k+1}+F_{3k}$. As both terms are even, their sum is also even.
Thus, the statement is also true for $n=k+1$
Step 5. Conclusion: with mathematical induction, we have proved that the statement is true for all natural numbers $n$.
