Answer
$a_{n}=\displaystyle \frac{3n+2}{n!}, \quad n=1,2,3...$
Work Step by Step
Observe numerators and denominators separately, as sequences. Annotate $a_{n}=\displaystyle \frac{A_{n}}{B_{n}}$
The differences between terms of $A_{n}$ are 3,3,3,3,...
These are constant, similar to the sequence of natural numbers:
1,2,3,4,... where the difference between terms is $1$.
If we multiply each term of (1,2,3,4,...) with 3, we get
$3,6,9,12,15,...$
(a sequence with the general n-th term = $3n$.)
The difference between terms is now $3$, as in the problem.
If we add 2 to each term of $(3,6,9,12,15,...)$,
we get the numerator sequence $A_{n}$ of our problem..
The general term for the numerators is, $A_{n}=3n+2$
We recognize the denominators from
$1!=1$
$2!=2$
$3!=2!\times 3=6$
$4!=3!\times 4=24$
$5!=4!\times 5=120,$
etc.
Thus, the denominators form the sequence for which $B_{n}=n!$
So,
$a_{n}=\displaystyle \frac{3n+2}{n!}, \quad n=1,2,3...$
