Use the Principle of Mathematical Induction to show that for a sequence $d_{1}, d_{2}, d_{3}, \ldots$ defined by $d_{1}=2$ and $d_{k}=\frac{d_{k-1}}{k}$ for all $k \geq 2$,the general term is $d_{n}=\frac{2}{n!}$ for all $n \in N$.

(N/A) Let $P(n)$ be the statement $d_{n} = \frac{2}{n!}$ for $n \in N$.
Step $1$: For $n=1$,$d_{1} = \frac{2}{1!} = \frac{2}{1} = 2$. This matches the given $d_{1}=2$. So,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N$,i.e.,$d_{k} = \frac{2}{k!}$.
Step $3$: We need to show $P(k+1)$ is true,i.e.,$d_{k+1} = \frac{2}{(k+1)!}$.
Given $d_{k+1} = \frac{d_{k}}{k+1}$.
Substituting the assumption $d_{k} = \frac{2}{k!}$,we get:
$d_{k+1} = \frac{2/k!}{k+1} = \frac{2}{k!(k+1)} = \frac{2}{(k+1)!}$.
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Conclusion: By the Principle of Mathematical Induction,$P(n)$ is true for all $n \in N$.