Question

In the following exercises, write the first five terms of the sequence whose general term is given.

$a_n=\frac{(n+2)!}{(n+3)!}$

Short answer

The first five terms of the sequence $a_n=\frac{(n+2)!}{(n+3)!}$is $\frac{1}{4},\frac{1}{5},\frac{1}{6},\frac{1}{7},\frac{1}{8}$

Step-by-step solution

Step 1. Given

The general term of the sequence is$a_n=\frac{(n+2)!}{(n+3)!}$

To find the first five terms of the sequence.

Step 2. Substitute 1,2 in the general form

Substitute the values $1,2$in the general form of the sequence,

$$\begin{gathered} a_1=\frac{\left(\right(1)+2)!}{\left(\right(1)+3)!} \\ =\frac{3!}{4!} \\ =\frac{1}{4} \end{gathered}$$

$$\begin{gathered} a_2=\frac{\left(\right(2)+2)!}{\left(\right(2)+3)!} \\ =\frac{4!}{5!} \\ =\frac{1}{5} \end{gathered}$$

Step 3. Substitute 3,4,5 in the general from.

$$\begin{gathered} a_3=\frac{\left(\right(3)+2)!}{\left(\right(3)+3)!} \\ =\frac{5!}{6!} \\ =\frac{1}{6} \end{gathered}$$

$$\begin{gathered} a_4=\frac{\left(\right(4)+2)!}{\left(\right(4)+3)!} \\ =\frac{6!}{7!} \\ =\frac{1}{7} \end{gathered}$$

$$\begin{gathered} a_5=\frac{\left(\right(5)+2)!}{\left(\right(5)+3)!} \\ =\frac{7!}{8!} \\ =\frac{1}{8} \end{gathered}$$

The five terms of the sequence are $\frac{1}{4},\frac{1}{5},\frac{1}{6},\frac{1}{7},\frac{1}{8}$