Question

In the following exercises, write each sum using summation notation.

$-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+\frac{1}{81}-\frac{1}{243}$

Short answer

The summation notation is$\sum _{n=1}^5{\left(-1\right)}^n\frac{1}{3^n}$.

Step-by-step solution

Step 1. Given information.

The given expression is,

$-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+\frac{1}{81}-\frac{1}{243}$.

Step 2. Finding the summation notation.

Here, the numerators are all one's and the denominators are $3$power from $1$to $5$with alternative signs then we get,

localid="1645358215087" $$\begin{gathered} -\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+\frac{1}{81}-\frac{1}{243} \\ n:1,2,3,4,5 \\ Terms:-\frac{1}{3},\frac{1}{9},-\frac{1}{27},\frac{1}{81},-\frac{1}{243} \\ Pattern:{\left(-1\right)}^1\frac{1}{3^1},{\left(-1\right)}^2\frac{1}{3^2},{\left(-1\right)}^3\frac{1}{3^3},{\left(-1\right)}^4\frac{1}{3^4},{\left(-1\right)}^5\frac{1}{3^5},........{\left(-1\right)}^n\frac{1}{3^n} \end{gathered}$$

Then the sum can be written in the form of summation notation that is

$-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+\frac{1}{81}-\frac{1}{243}=\sum _{n=1}^5{\left(-1\right)}^n\frac{1}{3^n}$.