Chapter 13.5, Problem 46E

To determine

To calculate: The sum of the expression ${\displaystyle \sum _{b=1}^{\infty }3{\left(0.5\right)}^b}$ and express it in expanded form.

Answer to Problem 46E

The expanded formis $3{\left(0.5\right)}^1+3{\left(0.5\right)}^2+3{\left(0.5\right)}^3+\cdots$ . The sum of the expression is 3.

Explanation of Solution

Given information:

The expression ${\displaystyle \sum _{b=1}^{\infty }3{\left(0.5\right)}^b}$ .

Formula used:

The sum of infinite terms of a geometric progression whose first term is a and common ratio ris $S=\frac{a}{1-r}$ .

Calculation:

Consider the expression ${\displaystyle \sum _{b=1}^{\infty }3{\left(0.5\right)}^b}$ .

Substitute 1, 2, 3 and so on values for b to express the expression in expanded form,

  ${\displaystyle \sum _{b=1}^{\infty }3{\left(0.5\right)}^b}=3{\left(0.5\right)}^1+3{\left(0.5\right)}^2+3{\left(0.5\right)}^3+\cdots$

The above terms $3{\left(0.5\right)}^1+3{\left(0.5\right)}^2+3{\left(0.5\right)}^3+\cdots$ is rewritten as,

  $3\left[{\left(0.5\right)}^1+{\left(0.5\right)}^2+{\left(0.5\right)}^3+\cdots \right]$

This forms a geometric progression with first term as $\left(0.5\right)$ and common ratio $\left(0.5\right)$ .

Recall that the sum of infinite terms of a geometric progression whose first term is a and common ratio ris $S=\frac{a}{1-r}$ .

Apply it,

  $\begin{array}{c}{\displaystyle \sum _{b=1}^{\infty }3{\left(0.5\right)}^b}=3{\left(0.5\right)}^1+3{\left(0.5\right)}^2+3{\left(0.5\right)}^3+\cdots \\ =3\left[{\left(0.5\right)}^1+{\left(0.5\right)}^2+{\left(0.5\right)}^3+\cdots \right]\\ =3\left(\frac{0.5}{1-0.5}\right)\\ =3\left(\frac{0.5}{0.5}\right)\end{array}$

Simplify it further as,

  $\begin{array}{c}{\displaystyle \sum _{b=1}^{\infty }3{\left(0.5\right)}^b}=3\left(\frac{0.5}{0.5}\right)\\ =3\end{array}$

Thus, the expanded form is $3{\left(0.5\right)}^1+3{\left(0.5\right)}^2+3{\left(0.5\right)}^3+\cdots$ . The sum of the expression is 3.