Prove the statement by the Principle of Mathematical Induction: $4^{n}-1$ is divisible by $3$,for each natural number $n$.

Let $P(n): 4^{n}-1$ is divisible by $3$ for all $n \in \mathbb{N}$.
Step $1$: For $n=1$,$P(1) = 4^{1}-1 = 3$,which is divisible by $3$. Thus,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in \mathbb{N}$,i.e.,$4^{k}-1 = 3m$ for some integer $m \in \mathbb{N}$. This implies $4^{k} = 3m+1$ $(i)$.
Step $3$: We need to prove $P(k+1)$ is true,i.e.,$4^{k+1}-1$ is divisible by $3$.
Consider $4^{k+1}-1 = 4 \cdot 4^{k}-1$.
Substituting from $(i)$,we get $4(3m+1)-1 = 12m+4-1 = 12m+3 = 3(4m+1)$.
Since $3(4m+1)$ is a multiple of $3$,$4^{k+1}-1$ is divisible by $3$.
Conclusion: By the Principle of Mathematical Induction,$P(n)$ is true for all $n \in \mathbb{N}$.