Solve for a2 a_2=(2^3)/(2!)

$a_{2} = \frac{2^{3}}{2!}$

Raise $2$ to the power of $3$ .

$a_{2} = \frac{8}{2!}$

Simplify the denominator.

Expand $2!$ to $2 \cdot 1$ .

$a_{2} = \frac{8}{2 \cdot 1}$

Multiply $2$ by $1$ .

$a_{2} = \frac{8}{2}$

Divide $8$ by $2$ .

$a_{2} = 4$