Question

Please solve the problem with steps

Answer 1

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:

$\frac{3^{n-1}-1}{6^{n-1}} =\frac{3^{n-1}}{6^{n-1}} -\frac{1}{6^{n-1}} =(\frac{1}{2})^{n-1} -\frac{1}{6^{n-1}}$

The first one: $(\frac{1}{2})^{n-1}$ is a geometric sequence of first term ($a_1$) "1" and common ratio (r) " $\frac{1}{2}$ ", so since the common ratio is smaller than one, we can find an answer for the infinite addition of its terms, given by: $Infinite\,Sum=\frac{a_1}{1-r} = \frac{1}{1-\frac{1}{2} } =\frac{1}{\frac{1}{2} } =2$

The second one: $\frac{1}{6^{n-1}}$ is a geometric sequence of first term "1", and common ratio (r) " $\frac{1}{6}$ ". Again, since the common ratio is smaller than one, we can find its infinite sum:

$Infinite\,Sum=\frac{a_1}{1-r} = \frac{1}{1-\frac{1}{6} } =\frac{1}{\frac{5}{6} } =\frac{6}{5}$

now we simply combine the results making sure we do the indicated difference: Infinite total sum= $2-\frac{6}{5} =\frac{10-6}{5} =\frac{4}{5}$

Answer 2

Answer:

HOPE YOU FOUND YOUR ANSWER

Step-by-step explanation: