Problem 39 - College Algebra

Question

Find the sum of each infinite geometric series. $$3+\frac{3}{4}+\frac{3}{4^{2}}+\frac{3}{4^{3}}+\cdots$$

Step-by-Step Solution

Answer

The sum of the infinite geometric series is 4.

Step 1: Identify the First Term and the Common Ratio

The first term \(a\) of the series is 3 and the common ratio \(r\) is \(\frac{3}{4}\). This can be seen because each succeeding term after 3 is obtained by multiplying the preceding term by \(\frac{1}{4}\), for example, \(\frac{3}{4}=\frac{3}{4^{1}}=3 \times \(\frac{1}{4}\), \(\frac{3}{4^{2}}=3 \times \(\frac{1}{4}\)^{2}\), and so on.

Step 2: Substitute the values into the Series Sum Formula

To compute the sum of the series, you substitute \(a = 3\) and \(r = \frac{1}{4}\) into the series sum formula \(S=\frac{a}{1-r}\). This gives: \(S=\frac{3}{1-\frac{1}{4}}\).

Step 3: Solve the Expression

Then you simplify the expression on the right-hand side of the equation to find the sum \(S\). The computation is as follows: \(S=\frac{3}{1-\frac{1}{4}} = \frac{3}{\frac{3}{4}} = \frac{3}{1} \times \frac{4}{3} = 4.