Use the Principle of Mathematical Induction to show that for a sequence $a_{1}, a_{2}, a_{3}, \ldots$ defined by $a_{1}=3$ and $a_{k}=7 a_{k-1}$ for all natural numbers $k > 1$,the general term is $a_{n}=3 \cdot 7^{n-1}$ for all $n \in N$.

(A) Let $P(n)$ be the statement $a_{n} = 3 \cdot 7^{n-1}$ for $n \in N$.
Step $1$: For $n=1$,$a_{1} = 3 \cdot 7^{1-1} = 3 \cdot 7^{0} = 3(1) = 3$. This matches the given $a_{1} = 3$. Thus,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N$,i.e.,$a_{k} = 3 \cdot 7^{k-1}$.
Step $3$: We need to show $P(k+1)$ is true,i.e.,$a_{k+1} = 3 \cdot 7^{(k+1)-1} = 3 \cdot 7^{k}$.
From the given recurrence relation,$a_{k+1} = 7 a_{k}$.
Substituting the assumption $a_{k} = 3 \cdot 7^{k-1}$,we get:
$a_{k+1} = 7 \cdot (3 \cdot 7^{k-1}) = 3 \cdot 7^{1} \cdot 7^{k-1} = 3 \cdot 7^{k}$.
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
By the Principle of Mathematical Induction,$a_{n} = 3 \cdot 7^{n-1}$ is true for all $n \in N$.