Chapter 9.1, Problem 91E

a

To determine

To find :The third partial sum of the series ${\displaystyle \sum _{n=1}^{\infty }4{\left(-\frac{1}{2}\right)}^n}$ .

Answer to Problem 91E

  $-\frac{3}{2}$

Explanation of Solution

Given:

  ${\displaystyle \sum _{n=1}^{\infty }4{\left(-\frac{1}{2}\right)}^n}$

Calculation:

Given series can be written as

  ${\displaystyle \sum _{n=1}^{\infty }4{\left(-\frac{1}{2}\right)}^n}=4\left(-\frac{1}{2}\right)+4{\left(-\frac{1}{2}\right)}^2+4{\left(-\frac{1}{2}\right)}^3+4{\left(-\frac{1}{2}\right)}^4+\mathrm{......}$

We know that nth partial sum of a series is the sum of first n terms of the series.

Therefore, the third partial sum of given series is

  $S_3=4\left(-\frac{1}{2}\right)+4{\left(-\frac{1}{2}\right)}^2+4{\left(-\frac{1}{2}\right)}^3$

which is a geometric series of 3 terms with first term $a=4\left(-\frac{1}{2}\right)=-2$ and common ratio $r=-\frac{1}{2}$

The sum of n of a geometric series with first term a and common ratio r is

  $S_n=\frac{a\left(1-r^n\right)}{1-r}$

Therefore, the third partial sum of given series is

  $S_3=\frac{-2\left(1-{\left(-\frac{1}{2}\right)}^3\right)}{1-\left(-\frac{1}{2}\right)}=\frac{-2\left(1+\frac{1}{8}\right)}{\frac{3}{2}}=-\frac{3}{2}$

b

To determine

To find :The fourth partial sum of the series ${\displaystyle \sum _{n=1}^{\infty }4{\left(-\frac{1}{2}\right)}^n}$ .

Answer to Problem 91E

  $-\frac{5}{4}$

Explanation of Solution

Given:

  ${\displaystyle \sum _{n=1}^{\infty }4{\left(-\frac{1}{2}\right)}^n}$

Calculation:

The fourth partial sum of given series is

  $\begin{array}{l}{S}_4=4\left(-\frac{1}{2}\right)+4{\left(-\frac{1}{2}\right)}^2+4{\left(-\frac{1}{2}\right)}^3+4{\left(-\frac{1}{2}\right)}^4\\ S_4=\frac{-2\left(1-{\left(-\frac{1}{2}\right)}^4\right)}{1-\left(-\frac{1}{2}\right)}=\frac{-2\left(1-\frac{1}{16}\right)}{\frac{3}{2}}=-\frac{5}{4}\end{array}$

c

To determine

To find :The fifth partial sum of the series ${\displaystyle \sum _{n=1}^{\infty }4{\left(-\frac{1}{2}\right)}^n}$ .

Answer to Problem 91E

  $-\frac{11}{8}$

Explanation of Solution

Given:

  ${\displaystyle \sum _{n=1}^{\infty }4{\left(-\frac{1}{2}\right)}^n}$

Calculation:

The fifth partial sum of given series is

  $\begin{array}{l}{S}_5=4\left(-\frac{1}{2}\right)+4{\left(-\frac{1}{2}\right)}^2+4{\left(-\frac{1}{2}\right)}^3+4{\left(-\frac{1}{2}\right)}^4+4{\left(-\frac{1}{2}\right)}^5\\ S_5=\frac{-2\left(1-{\left(-\frac{1}{2}\right)}^5\right)}{1-\left(-\frac{1}{2}\right)}=\frac{-2\left(1+\frac{1}{32}\right)}{\frac{3}{2}}=-\frac{11}{8}\end{array}$