Answer
$2,\dfrac{3}{2},\dfrac{8}{3},\dfrac{30}{4}$
Work Step by Step
Need to find the first four terms of $a_n=\dfrac{(n+1)!}{n^2}$ when $n=1,2,3,4$
Thus,
$a_1=\dfrac{(1+1)!}{1^2}=2$
$a_2=\dfrac{(2+1)!}{2^2}=\dfrac{3}{2}$
$a_3=\dfrac{(3+1)!}{3^2}=\dfrac{4 \cdot 3 \cdot 2 \cdot 1}{9}=\dfrac{8}{3}$
$a_4=\dfrac{(4+1)!}{4^2}=\dfrac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{16}=\dfrac{30}{4}$
Hence, the first four terms are: $2,\dfrac{3}{2},\dfrac{8}{3},\dfrac{30}{4}$
