Answer:
Infinite series equals 4/5
Step-by-step explanation:
Notice that the series can be written as a combination of two geometric series, that can be found independently:
[tex]\frac{3^{n-1}-1}{6^{n-1}} =\frac{3^{n-1}}{6^{n-1}} -\frac{1}{6^{n-1}} =(\frac{1}{2})^{n-1} -\frac{1}{6^{n-1}}[/tex]
The first one: [tex](\frac{1}{2})^{n-1}[/tex] is a geometric sequence of first term ([tex]a_1[/tex]) "1" and common ratio (r) " [tex]\frac{1}{2}[/tex] ", so since the common ratio is smaller than one, we can find an answer for the infinite addition of its terms, given by: [tex]Infinite\,Sum=\frac{a_1}{1-r} = \frac{1}{1-\frac{1}{2} } =\frac{1}{\frac{1}{2} } =2[/tex]
The second one: [tex]\frac{1}{6^{n-1}}[/tex] is a geometric sequence of first term "1", and common ratio (r) " [tex]\frac{1}{6}[/tex] ". Again, since the common ratio is smaller than one, we can find its infinite sum:
[tex]Infinite\,Sum=\frac{a_1}{1-r} = \frac{1}{1-\frac{1}{6} } =\frac{1}{\frac{5}{6} } =\frac{6}{5}[/tex]
now we simply combine the results making sure we do the indicated difference: Infinite total sum= [tex]2-\frac{6}{5} =\frac{10-6}{5} =\frac{4}{5}[/tex]
