Answer
1,2, $\frac{3}{2}$ , $\frac{2}{3}$
Work Step by Step
To find the first four terms of the sequence whose general term is $a_{n}$ =$\frac{n^{2}}{n!}$, we replace n in the formula with 1,2,3, and 4.
n=1, $a_{1}$ = $\frac{1^{2}}{1!}$ = 1
n=2, $a_{2}$ =$\frac{2^{2}}{2!}$ = $\frac{2*2}{2*1}$ = 2
n=3, $a_{3}$ =$\frac{3^{2}}{3!}$ = $\frac{3*3}{3*2*1}$ = $\frac{3}{2}$
n=4,$a_{4}$ =$\frac{4^{2}}{4!}$ = $\frac{4*4}{4*3*2*1}$ = $\frac{2}{3}$
The first four terms are 1,2, $\frac{3}{2}$ , $\frac{2}{3}$.
