Question

Write the first five terms of the sequence whose general term is$a_n=\frac{3}{n!}$

Short answer

The first five terms of the sequence are$3,\frac{3}{2},\frac{1}{2},\frac{3}{8}\text{and}\frac{1}{40}.$

Step-by-step solution

Step 1. Given information

$\begin{array}{l}\text{Thegivengeneraltermis}\\ a_n=\frac{3}{n!}\\ \text{Wehavetofindthefirstfivetermofthesequence.}\end{array}$

Step 2. Calculating the first five terms 

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

localid="1645105874281" $\begin{array}{l}\text{Weknow}\\ \text{If}n\text{isapositiveinteger,then}\\ \Rightarrow n!=n\left(n-1\right)\left(n-2\right)\dots \\ \text{Now,}\\ \begin{array}{ccccc}n=1& n=2& n=3& n=4& n=5\\ \begin{array}{l}\Rightarrow a_1=\frac{3}{1!}\\ =\frac{3}{1}\\ =3\end{array}& \begin{array}{l}\Rightarrow a_2=\frac{3}{2!}\\ =\frac{3}{2\cdot 1}\\ =\frac{3}{2}\end{array}& \begin{array}{l}\Rightarrow a_3=\frac{3}{3!}\\ =\frac{3}{3\cdot 2\cdot 1}\\ =\frac{1}{2}\end{array}& \begin{array}{l}\Rightarrow a_4=\frac{3}{4!}\\ =\frac{3}{4\cdot 3\cdot 2\cdot 1}\\ =\frac{3}{8}\end{array}& \begin{array}{l}\Rightarrow a_5=\frac{3}{5!}\\ =\frac{3}{5\cdot 4\cdot 3\cdot 2\cdot 1}\\ =\frac{1}{40}\end{array}\end{array}\end{array}$

Therefore, the first five terms of the sequence will be$3,\frac{3}{2},\frac{1}{2},\frac{3}{8}\text{and}\frac{1}{40}.$