Answer
See explanations.
Work Step by Step
Step 1. Recall the properties of Fibonacci numbers: $F_n=F_{n-1}+F_{n-2}$ and $F_1=F_2=1$
Step 2. Prove that the statement is true for $n=1$: $F_4=F_3+F_2=2F_2+F_1=3$ is divisible by 3, thus it is true for $n=1$
Step 3. Assume the statement is true for $n=k$: we have $F_{4k}=3m$ is divisible by 3, where m is an integer.
Step 4. Prove that it is also true for $n=k+1$:
$F_{4k+4}=F_{4k+3}+F_{4k+2}=F_{4k+2}+F_{4k+1}+F_{4k+1}+F_{4k}=F_{4k+1}+F_{4k}+2F_{4k+1}+F_{4k}
=3F_{4k+1}+2F_{4k}=3F_{4k+1}+6m=3(F_{4k+1}+2m)$ is divisible by 3.
Thus, the statement is also true for $n=k+1$
Step 5. With mathematical induction, we have proved that the statement is true for all natural numbers n.
