Chapter 9.1, Problem 92E

a

To determine

To find : The third partial sum of the series ${\displaystyle \sum _{n=1}^{\infty }5{\left(-\frac{1}{4}\right)}^n}$ .

Answer to Problem 92E

  $-\frac{65}{64}$

Explanation of Solution

Given:

  ${\displaystyle \sum _{n=1}^{\infty }5{\left(-\frac{1}{4}\right)}^n}$

Calculation:

Given series can be written as

  ${\displaystyle \sum _{n=1}^{\infty }5{\left(-\frac{1}{4}\right)}^n}=5\left(-\frac{1}{4}\right)+5{\left(-\frac{1}{4}\right)}^2+5{\left(-\frac{1}{4}\right)}^3+5{\left(-\frac{1}{4}\right)}^4+\mathrm{......}$

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

  $S_3=5\left(-\frac{1}{4}\right)+5{\left(-\frac{1}{4}\right)}^2+5{\left(-\frac{1}{4}\right)}^3$

which is a geometric series of 3 terms with first term $a=5\left(-\frac{1}{4}\right)=-\frac{5}{4}$ and common ratio $r=-\frac{1}{4}$

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

  $S_3=\frac{-\frac{5}{4}\left(1-{\left(-\frac{1}{4}\right)}^3\right)}{1-\left(-\frac{1}{4}\right)}=\frac{-\frac{5}{4}\left(1+\frac{1}{64}\right)}{\frac{5}{4}}=-\frac{65}{64}$

b

To determine

To find: The fourth partial sum of the series ${\displaystyle \sum _{n=1}^{\infty }5{\left(-\frac{1}{4}\right)}^n}$ .

Answer to Problem 92E

  $-\frac{255}{256}$

Explanation of Solution

Given:

  ${\displaystyle \sum _{n=1}^{\infty }5{\left(-\frac{1}{4}\right)}^n}$

Calculation:

The fourth partial sum of given series is

  $\begin{array}{l}{S}_4=5\left(-\frac{1}{4}\right)+5{\left(-\frac{1}{4}\right)}^2+5{\left(-\frac{1}{4}\right)}^3+5{\left(-\frac{1}{4}\right)}^4\\ S_4=\frac{-\frac{5}{4}\left(1-{\left(-\frac{1}{4}\right)}^4\right)}{1-\left(-\frac{1}{4}\right)}=\frac{-\frac{5}{4}\left(1-\frac{1}{256}\right)}{\frac{5}{4}}=-\frac{255}{256}\end{array}$

c

To determine

To find: The fifth partial sum of the series ${\displaystyle \sum _{n=1}^{\infty }5{\left(-\frac{1}{4}\right)}^n}$ .

Answer to Problem 92E

  $-\frac{1025}{1024}$

Explanation of Solution

Given:

  ${\displaystyle \sum _{n=1}^{\infty }5{\left(-\frac{1}{4}\right)}^n}$

Calculation:

The fifth partial sum of given series is

  $\begin{array}{l}{S}_5=5\left(-\frac{1}{4}\right)+5{\left(-\frac{1}{4}\right)}^2+5{\left(-\frac{1}{4}\right)}^3+5{\left(-\frac{1}{4}\right)}^4+5{\left(-\frac{1}{4}\right)}^5\\ S_5=\frac{-\frac{5}{4}\left(1-{\left(-\frac{1}{4}\right)}^5\right)}{1-\left(-\frac{1}{4}\right)}=\frac{-\frac{5}{4}\left(1+\frac{1}{1024}\right)}{\frac{5}{4}}=-\frac{1025}{1024}\end{array}$