Find the Second Term a_n=((n+4)!)/((n+5)!)

$a_{n} = \frac{(n + 4)!}{(n + 5)!}$

Step 1

Substitute in the value of $n$ to find the $n$ th term.

$a_{2} = \frac{((2) + 4)!}{((2) + 5)!}$

Step 2

Simplify the factorials.

Add $2$ and $4$ .

$a_{2} = \frac{(6)!}{((2) + 5)!}$

Add $2$ and $5$ .

$a_{2} = \frac{(6)!}{(7)!}$

Step 3

Rewrite $(7)!$ as $7 \cdot 6!$ .

$a_{2} = \frac{(6)!}{7 \cdot 6!}$

Step 4

Cancel the common factor of $(6)!$ .

Cancel the common factor.

$a_{2} = \frac{(6)!}{7 \cdot 6!}$

Rewrite the expression.

$a_{2} = \frac{1}{7}$