In this case, we'll have to carry out several steps to find the solution.
Step 01:
Data:
an=(n+3)!/(n+4)!
Step 02:
nth term sequence:
1. n = 1
[tex]a_1=\frac{(1+3)!}{(1+4)!}=\frac{4!}{5!}=\frac{4\cdot3\cdot2\cdot1}{5\cdot4\cdot3\cdot2\cdot1}=\frac{1}{5}[/tex]
2. n = 2
[tex]a_2=\frac{(2+3)!}{(2+4)!}=\frac{5!}{6!}=\frac{5\cdot4\cdot3\cdot2\cdot1}{6\cdot5\cdot4\cdot3\cdot2\cdot1}=\frac{1}{6}[/tex]
3. n = 3
[tex]a_3=\frac{(3+3)!}{(3+4)!}=\frac{6!}{7!}=\frac{6\cdot5\cdot4\cdot3\cdot2\cdot1}{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}=\frac{1}{7}[/tex]
4. n = 4
[tex]a_4=\frac{(4+3)!}{(4+4)!}=\frac{7!}{8!}=\frac{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}=\frac{1}{8}[/tex]
The answer is:
[tex]\frac{1}{5},\frac{1}{6},\frac{1}{7},\frac{1}{8}[/tex]
