Chapter 8.3, Problem 57E

To determine

To find: the sum and verify the result by graphing utility.

Answer to Problem 57E

  ${\displaystyle \sum _{i=1}^{7}64{\left(-\frac{1}{2}\right)}^{i-1}}=43$

Explanation of Solution

Given information:

Given sum

  ${\displaystyle \sum _{i=1}^{7}64{\left(-\frac{1}{2}\right)}^{i-1}}$

Formula used:

The sum $S_n$ or ${\displaystyle \sum _{i=1}^n{a}_r{r}^{r-1}}$ of a finite geometric sequence is $a_1\left(\frac{1-r^n}{1-r}\right)$ , where $a_1$ is the first term, $r$ is the common ration, and $n$ is the number of terms in the sequence.

Calculation:

Write out a few terms of the finite sequence.

  ${\displaystyle \sum _{i=1}^{7}64{\left(-\frac{1}{2}\right)}^{i-1}}$

  ${\displaystyle \sum _{i=1}^{7}64{\left(-\frac{1}{2}\right)}^{i-1}}=64\cdot {\left(-\frac{1}{2}\right)}^0+64\cdot {\left(-\frac{1}{2}\right)}^1+64\cdot {\left(-\frac{1}{2}\right)}^2+\mathrm{...}+64\cdot {\left(-\frac{1}{2}\right)}^6$

Substitute 64for $a_1$ , $\left(-\frac{1}{2}\right)$ for $r$ , and 7 for $n$ is $S_n=a_1\left(\frac{1-r^n}{1-r}\right)$ .

  $\begin{array}{c}{S}_7=64\left(\frac{1-{\left(-\frac{1}{2}\right)}^7}{1-\left(-\frac{1}{2}\right)}\right)\\ =\left(\frac{2^7+1}{2+1}\right)\\ =\frac{1}{3}\left(128+1\right)\\ =\frac{129}{3}\\ =43\end{array}$

Thus, the sum is 43.

Verify the result using graphing utility:

Verify the result using graphing utility as shown:

1. Choose the sum feature from the math menu of the list feature.
2. Choose the sequence feature from the operations menu of the list feature.
3. Enter the expression for the sequence, the variable, the lower limit of summation, and the upper limit of summation. Press ENTER to get the sum as 43.