What is the following value when expressed as a common fraction: $$\frac{1}{3^{1}}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\frac{1}{3^{4}}+\frac{1}{3^{5}}+\frac{1}{3^{6}}?$$
This is a finite geometric series with first term $\frac{1}{3}$, common ratio $\frac{1}{3}$ and $6$ terms. Therefore the sum is: $$\frac{\frac{1}{3}\left(1-\frac{1}{3^{6}}\right)}{1-\frac{1}{3}}
=\frac{\frac{3^{6}-1}{3^{7}}}{\frac{2}{3}}
= \frac{3^{6}-1}{2\cdot3^{6}}=\frac{729-1}{2\cdot 729} = \boxed{\frac{364}{729}}.$$
\frac{364}{729}
Level 4
Algebra
Problem: What is the following value when expressed as a common fraction: $$\frac{1}{3^{1}}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\frac{1}{3^{4}}+\frac{1}{3^{5}}+\frac{1}{3^{6}}?$$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the sum of the series, recognize that it is a geometric series. The series can be expressed as:

S = \(\frac{1}{3^{1}} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + \frac{1}{3^{4}} + \frac{1}{3^{5}} + \frac{1}{3^{6}}\).

Identify the first term (a) and the common ratio (r). Here, a = \(\frac{1}{3}\) and r = \(\frac{1}{3}\).

The formula for the sum of the first n terms of a geometric series is:

S_n = a \(\frac{1 - r^n}{1 - r}\).

In this case, n = 6. Substitute the values into the formula:

S_6 = \(\frac{1}{3} \cdot \frac{1 - (\frac{1}{3})^6}{1 - \frac{1}{3}}\).

Calculate the denominator:

1 - \(\frac{1}{3} = \frac{2}{3}\).

Now, substitute this back into the equation:

S_6 = \(\frac{1}{3} \cdot \frac{1 - \frac{1}{729}}{\frac{2}{3}}\).

Simplify the fraction:

S_6 = \(\frac{1}{3} \cdot \frac{3}{2} \cdot (1 - \frac{1}{729})\).

This simplifies to:

S_6 = \(\frac{1}{2} \cdot (1 - \frac{1}{729})\).

Calculate \(1 - \frac{1}{729}\):

1 - \(\frac{1}{729} = \frac{728}{729}\).

Now substitute this back into the equation:

S_6 = \(\frac{1}{2} \cdot \frac{728}{729}\).

Multiply the fractions:

S_6 = \(\frac{728}{1458}\).

Now simplify \(\frac{728}{1458}\). The greatest common divisor of 728 and 1458 is 2:

S_6 = \(\frac{364}{729}\).

Thus, the final answer is \(\frac{364}{729}\).<|END_OF_LATENT|> The final answer is $\frac{364}{729}$.