Question

Write the first five terms of the sequence whose general term is$a_n=\frac{2}{n!}.$

Short answer

The first five terms of the given sequence are$2,1,\frac{1}{3},\frac{1}{12}and\frac{1}{60}.$

Step-by-step solution

Step 1. Given information

$\begin{array}{l}\text{Thegivengeneraltermis}\\ a_n=\frac{2}{n!}\\ \text{Wehavetofindthefirstfivetermofthesequence.}\end{array}$

Step 2. Calculating the first five terms

$\begin{array}{l}\text{Weknow}\\ \text{If}n\text{isapositiveinteger,then}\\ \Rightarrow n!=n\left(n-1\right)\left(n-2\right)\dots \end{array}$

We substitute the values 1, 2, 3, 4, 5 into the formula $a_n=\frac{2}{n!}$localid="1645104944104" $\begin{array}{ccccc}n=1& n=2& n=3& n=4& n=5\\ \begin{array}{l}\Rightarrow a_1=\frac{2}{1!}\\ =\frac{2}{1}\\ =2\end{array}& \begin{array}{l}\Rightarrow a_2=\frac{2}{2!}\\ =\frac{2}{2\cdot 1}\\ =1\end{array}& \begin{array}{l}\Rightarrow a_3=\frac{2}{3!}\\ =\frac{2}{3\cdot 2\cdot 1}\\ =\frac{1}{3}\end{array}& \begin{array}{l}\Rightarrow a_4=\frac{2}{4!}\\ =\frac{2}{4\cdot 3\cdot 2\cdot 1}\\ =\frac{1}{12}\end{array}& \begin{array}{l}\Rightarrow a_5=\frac{2}{5!}\\ =\frac{2}{5\cdot 4\cdot 3\cdot 2\cdot 1}\\ =\frac{1}{60}\end{array}\end{array}$

Therefore, the first five terms will be$2,1,\frac{1}{3},\frac{1}{12}and\frac{1}{60}.$