Evaluate:$\[ \sum_{n=1}^6 \frac{1}{6}(4)^{n-1} \\]What Is The Value Of \[$ S \$\]?

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Introduction

In mathematics, summation is a fundamental concept used to represent the sum of a series of numbers. It is denoted by the symbol $\sum$ and is used to calculate the total value of a sequence of numbers. In this article, we will evaluate the summation $\sum_{n=1}^6 \frac{1}{6}(4)^{n-1}$ and find its value.

Understanding the Summation

The given summation is a finite series, which means it has a finite number of terms. The series is defined as:

$\sum_{n=1}^6 \frac{1}{6}(4)^{n-1}$

This can be expanded as:

$\frac{1}{6}(4)^0 + \frac{1}{6}(4)^1 + \frac{1}{6}(4)^2 + \frac{1}{6}(4)^3 + \frac{1}{6}(4)^4 + \frac{1}{6}(4)^5$

Evaluating the Series

To evaluate the series, we can start by calculating each term separately.

$\frac{1}{6}(4)^0 = \frac{1}{6}(1) = \frac{1}{6}$

$\frac{1}{6}(4)^1 = \frac{1}{6}(4) = \frac{2}{3}$

$\frac{1}{6}(4)^2 = \frac{1}{6}(16) = \frac{8}{3}$

$\frac{1}{6}(4)^3 = \frac{1}{6}(64) = \frac{32}{3}$

$\frac{1}{6}(4)^4 = \frac{1}{6}(256) = \frac{128}{3}$

$\frac{1}{6}(4)^5 = \frac{1}{6}(1024) = \frac{512}{3}$

Finding the Sum

Now that we have calculated each term, we can find the sum of the series by adding all the terms together.

$S = \frac{1}{6} + \frac{2}{3} + \frac{8}{3} + \frac{32}{3} + \frac{128}{3} + \frac{512}{3}$

To add these fractions, we need to find a common denominator, which is 6 in this case.

$S = \frac{1}{6} + \frac{4}{6} + \frac{16}{6} + \frac{64}{6} + \frac{256}{6} + \frac{1024}{6}$

Now we can add the numerators:

$S = \frac{1 + 4 + 16 + 64 + 256 + 1024}{6}$

$S = \frac{1365}{6}$

Simplifying the Fraction

To simplify the fraction, we can divide the numerator and denominator by their greatest common divisor, which is 3 in this case.

$S = \frac{1365 \div 3}{6 \div 3}$

$S = \frac{455}{2}$

Conclusion

In this article, we evaluated the summation $\sum_{n=1}^6 \frac{1}{6}(4)^{n-1}$ and found its value to be $\frac{455}{2}$. This is a finite series, and we calculated each term separately before finding the sum. The result is a simplified fraction, which is the final answer to the problem.

Final Answer

The final answer to the problem is $\boxed{\frac{455}{2}}$.

Discussion

The given summation is a geometric series, which is a series of the form:

$a + ar + ar^2 + ar^3 + ... + ar^{n-1}$

where $a$ is the first term and $r$ is the common ratio. In this case, the first term is $\frac{1}{6}$ and the common ratio is 4.

The formula for the sum of a geometric series is:

$S = \frac{a(1 - r^n)}{1 - r}$

where $n$ is the number of terms. In this case, $n = 6$, so we can plug in the values to get:

$S = \frac{\frac{1}{6}(1 - 4^6)}{1 - 4}$

$S = \frac{\frac{1}{6}(1 - 4096)}{-3}$

$S = \frac{\frac{1}{6}(-4095)}{-3}$

$S = \frac{455}{2}$

This is the same result we got by evaluating the series term by term.

Applications

Geometric series have many applications in mathematics and other fields. Some examples include:

• Finance: Geometric series are used to calculate the future value of an investment or loan.
• Physics: Geometric series are used to calculate the total energy of a system.
• Computer Science: Geometric series are used to calculate the time complexity of algorithms.

References

• "Geometric Series" by Math Open Reference
• "Sum of a Geometric Series" by Wolfram MathWorld
• "Geometric Series Formula" by Khan Academy

Note: The references provided are for general information and are not specific to the problem at hand.

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Introduction

In our previous article, we evaluated the summation $\sum_{n=1}^6 \frac{1}{6}(4)^{n-1}$ and found its value to be $\frac{455}{2}$. In this article, we will answer some frequently asked questions related to this problem.

Q1: What is the formula for the sum of a geometric series?

A1: The formula for the sum of a geometric series is:

$S = \frac{a(1 - r^n)}{1 - r}$

where $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms.

Q2: How do I calculate the sum of a geometric series?

A2: To calculate the sum of a geometric series, you can use the formula above. First, identify the first term, common ratio, and number of terms. Then, plug these values into the formula and simplify.

Q3: What is the difference between a geometric series and an arithmetic series?

A3: A geometric series is a series of the form:

$a + ar + ar^2 + ar^3 + ... + ar^{n-1}$

where $a$ is the first term and $r$ is the common ratio. An arithmetic series, on the other hand, is a series of the form:

$a + (a + d) + (a + 2d) + (a + 3d) + ... + (a + (n-1)d)$

where $a$ is the first term and $d$ is the common difference.

Q4: How do I find the sum of a finite geometric series?

A4: To find the sum of a finite geometric series, you can use the formula:

$S = \frac{a(1 - r^n)}{1 - r}$

where $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms.

Q5: What is the significance of the common ratio in a geometric series?

A5: The common ratio in a geometric series determines the rate at which the terms increase or decrease. If the common ratio is greater than 1, the terms increase exponentially. If the common ratio is less than 1, the terms decrease exponentially.

Q6: How do I determine the number of terms in a geometric series?

A6: To determine the number of terms in a geometric series, you can use the formula:

$n = \log_r \left( \frac{S - a}{a(r - 1)} \right)$

where $S$ is the sum of the series, $a$ is the first term, and $r$ is the common ratio.

Q7: What is the relationship between the sum of a geometric series and the first term?

A7: The sum of a geometric series is directly proportional to the first term. If the first term is increased or decreased, the sum of the series will also increase or decrease proportionally.

Q8: How do I find the sum of an infinite geometric series?

A8: To find the sum of an infinite geometric series, you can use the formula:

$S = \frac{a}{1 - r}$

where $a$ is the first term and $r$ is the common ratio.

Q9: What is the significance of the first term in a geometric series?

A9: The first term in a geometric series determines the starting value of the series. If the first term is increased or decreased, the entire series will shift up or down accordingly.

Q10: How do I determine the common ratio in a geometric series?

A10: To determine the common ratio in a geometric series, you can use the formula:

$r = \frac{a_{n+1}}{a_n}$

where $a_{n+1}$ is the next term in the series and $a_n$ is the current term.

Conclusion

In this article, we answered some frequently asked questions related to the summation $\sum_{n=1}^6 \frac{1}{6}(4)^{n-1}$. We hope that this Q&A article has provided you with a better understanding of geometric series and how to evaluate them.

Final Answer

The final answer to the problem is $\boxed{\frac{455}{2}}$.

Discussion

Geometric series have many applications in mathematics and other fields. Some examples include:

• Finance: Geometric series are used to calculate the future value of an investment or loan.
• Physics: Geometric series are used to calculate the total energy of a system.
• Computer Science: Geometric series are used to calculate the time complexity of algorithms.

References

• "Geometric Series" by Math Open Reference
• "Sum of a Geometric Series" by Wolfram MathWorld
• "Geometric Series Formula" by Khan Academy