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

Short Answer

Expert verified
The sum of the infinite geometric series is 4.

Step by step solution

01

Identify the first term and the ratio

In the series \(3+\frac{3}{4}+\frac{3}{4^{2}}+\frac{3}{4^{3}}+\cdots\), the first term 'a' is 3. The ratio 'r' is the quotient of the second term and the first, that is, \(\frac{3/4}{3} = \frac{1}{4}\).
02

Apply the Infinite Geometric Series Sum Formula

To find the sum of an infinite geometric series, we use the formula \(S = \frac{a}{1-r}\). Substituting \(a = 3\) and \(r = 1/4\) into the formula, we have \(S = \frac{3}{1 - \frac{1}{4}}\).
03

Simplify the Expression

Simplify the denominator by subtracting \(\frac{1}{4}\) from 1 and then divide 3 by this result. So, \(S = \frac{3}{\frac{3}{4}} = 4.\)

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