Chapter 11.4, Problem 23WE

To determine

To calculate:

The following series with sigma notation:

  $6-12+24-48$

Answer to Problem 23WE

The sigma notation for $6-12+24-48$ is ${\displaystyle \sum _{n=0}^{3}6{\left(2\right)}^n}$ .

Explanation of Solution

Given information:

  $6-12+24-48$

Calculation:

In the given series:

The first term is $6$ and the last term is $-48$ . Here, each term appears as $6$ times $2$ to the power of the index. Therefore the summand includes $6{\left(2\right)}^n$

The value of the upper limit is calculated by using the last term as:

  $\begin{array}{l}6{\left(-2\right)}^n=-48\\ {\left(-2\right)}^n=-8\\ {\left(-2\right)}^n={\left(-2\right)}^3\\ n=3\end{array}$

The value of the lower limit is calculated by using the last term as:

  $\begin{array}{l}6{\left(-2\right)}^n=6\\ {\left(-2\right)}^n=1\\ {\left(-2\right)}^n={\left(1\right)}^0\\ n=0\end{array}$

So, the lower and upper limit is $0$ and $3$ . Now, the sigma notation for the given series is expressed as:

  $6-12+24-48={\displaystyle \sum _{n=0}^{3}6{\left(2\right)}^n}$