Chapter 12, Problem 4CT

To determine

To find: The sum of the sequence.

Answer to Problem 4CT

The evaluated sum is $=-\frac{680}{81}$ .

Explanation of Solution

Given information:

The given terms are ${\displaystyle \sum _{k=1}^4\left[{\left(\frac{2}{3}\right)}^k-k\right]}$ .

Calculation:

Substitute $1$ for $k$ in the sequence to get the first term.

  $\begin{array}{c}{a}_1={\left(\frac{2}{3}\right)}^1-1\\ =-\frac{1}{3}\end{array}$

Similarly, find the second, third and fourth terms.

  $\begin{array}{c}{a}_2={\left(\frac{2}{3}\right)}^2-2\\ =\frac{4}{9}-2\\ =-\frac{14}{9}\end{array}$

  $\begin{array}{c}{a}_3={\left(\frac{2}{3}\right)}^3-3\\ =\frac{8}{27}-3\\ =-\frac{73}{27}\end{array}$

  $\begin{array}{c}{a}_4={\left(\frac{2}{3}\right)}^4-4\\ =\frac{16}{81}-4\\ =-\frac{308}{81}\end{array}$

In order to find the sum, add the first four terms.

  $\begin{array}{c}{\displaystyle \sum _{k=1}^4\left[{\left(\frac{2}{3}\right)}^k-k\right]}=-\frac{1}{3}+\left(-\frac{14}{9}\right)+\left(-\frac{73}{27}\right)+\left(-\frac{308}{81}\right)\\ =-\frac{1}{3}-\frac{14}{9}-\frac{73}{27}-\frac{308}{81}\\ =\frac{-27-126-219-308}{81}\\ =-\frac{680}{81}\end{array}$

Therefore, the evaluate sum is $=-\frac{680}{81}$ .