Chapter B, Problem 30AYU

In Problems 29-32, find the first five terms in the sequence of partial sums for each series.

${\displaystyle \underset{k=1}{\overset{\infty }{\sum }}{\left(\frac{1}{4}\right)}^{k-1}}$

To determine

To find: The first five terms in the sequence of partial sums for each series.

Answer to Problem 30AYU

Solution:

The first five terms in the sequence of partial sums for each series is $1, \left(\frac{5}{4}\right), \left(\frac{21}{16}\right), \left(\frac{85}{64}\right), \left(\frac{341}{256}\right)$.

Explanation of Solution

Given:

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

Calculation:

$\begin{array}{l}{\displaystyle {\sum }_{k = 1}^{\infty }{\left(\frac{1}{4}\right)}^{k - 1} = {\left(\frac{1}{4}\right)}^{1 - 1} + {\left(\frac{1}{4}\right)}^{2 - 1} + {\left(\frac{1}{4}\right)}^{3 - 1} + {\left(\frac{1}{4}\right)}^{4 - 1} + {\left(\frac{1}{4}\right)}^{5 - 1} + {\left(\frac{1}{4}\right)}^{6 - 1} + {\left(\frac{1}{4}\right)}^{7 - 1}\dots \dots \dots \dots \dots \dots }\\ = 1 + \left(\frac{1}{4}\right) + {\left(\frac{1}{4}\right)}^2 + {\left(\frac{1}{4}\right)}^3 + {\left(\frac{1}{4}\right)}^4 + {\left(\frac{1}{4}\right)}^5 + {\left(\frac{1}{4}\right)}^6 + \cdots \dots \dots \dots \dots \dots \\ = 1 + \left(\frac{1}{4}\right) + \left(\frac{1}{16}\right) + \left(\frac{1}{64}\right) + \left(\frac{1}{256}\right) + {\left(\frac{1}{4}\right)}^5 + {\left(\frac{1}{4}\right)}^6 + \cdots \dots \dots \dots \dots \dots \\ = 1 + \left(\frac{1}{4}\right) + \left(\frac{1}{16}\right) + \left(\frac{1}{64}\right) + \left(\frac{1}{256}\right) + \cdots \dots \mathrm{..}\end{array}$

The first five terms in the sequence of partial sums for each series is

$\begin{array}{l}1\\ 1 + \left(\frac{1}{4}\right) = \left(\frac{5}{4}\right)\\ 1 + \left(\frac{1}{4}\right) + \left(\frac{1}{16}\right) = \left(\frac{21}{16}\right)\\ 1 + \left(\frac{1}{4}\right) + \left(\frac{1}{16}\right) + \left(\frac{1}{64}\right) = \left(\frac{85}{64}\right)\\ 1 + \left(\frac{1}{4}\right) + \left(\frac{1}{16}\right) + \left(\frac{1}{64}\right) + \left(\frac{1}{256}\right) = \left(\frac{341}{256}\right)\end{array}$

The first five terms in the sequence of partial sums for each series is $1, \left(\frac{5}{4}\right), \left(\frac{21}{16}\right), \left(\frac{85}{64}\right), \left(\frac{341}{256}\right)$.