Question

Write the first five terms of the sequence whose general term is$a_n=\frac{(n-1)!}{(n+1)!}$

Short answer

The first five terms of the sequence with the general term as$a_n=\frac{(n-1)!}{(n+1)!}$are$\frac{1}{2},\frac{1}{6},\frac{1}{12},\frac{1}{20}\text{and}\frac{1}{30}.$

Step-by-step solution

Step 1. Given information

$\begin{array}{l}\text{Thegeneraltermisgivenas}\\ a_n=\frac{(n-1)!}{(n+1)!}\\ \text{Wehavetowritethefirst5termsofthesequence.}\end{array}$

Step 2. Calculating the first five terms 

We substitute the values 1, 2, 3, 4, 5 into the formula $a_n=\frac{(n-1)!}{(n+1)!}$

$\begin{array}{l}\text{Weknow}\\ \text{If}n\text{isapositiveinteger,then}\\ \Rightarrow n!=n\left(n-1\right)\left(n-2\right)\dots \\ \text{Now,}\\ \begin{array}{ccccc}n=1& n=2& n=3& n=4& n=5\\ \begin{array}{l}\Rightarrow a_1=\frac{(1-1)!}{(1+1)!}\\ =\frac{0!}{2!}\\ =\frac{1}{2\times 1}\\ =\frac{1}{2}\end{array}& \begin{array}{l}\Rightarrow a_2=\frac{(2-1)!}{(2+1)!}\\ =\frac{1!}{3!}\\ =\frac{1}{3\times 2\times 1}\\ =\frac{1}{6}\end{array}& \begin{array}{l}\Rightarrow a_3=\frac{(3-1)!}{(3+1)!}\\ =\frac{2!}{4!}\\ =\frac{2\times 1}{4\times 3\times 2\times 1}\\ =\frac{1}{12}\end{array}& \begin{array}{l}\Rightarrow a_4=\frac{(4-1)!}{(4+1)!}\\ =\frac{3!}{5!}\\ =\frac{3\times 2\times 1}{5\times 4\times 3\times 2\times 1}\\ =\frac{1}{20}\end{array}& \begin{array}{l}\Rightarrow a_5=\frac{(5-1)!}{(5+1)!}\\ =\frac{4!}{6!}\\ =\frac{4\times 3\times 2\times 1}{6\times 5\times 4\times 3\times 2\times 1}\\ =\frac{1}{30}\end{array}\end{array}\end{array}$

Step 3. Conclusion

From the above step, we can observe that the first five terms of the sequence will be$\frac{1}{2},\frac{1}{6},\frac{1}{12},\frac{1}{20}\text{and}\frac{1}{30}.$