Calculate The Value Of { N $}$ If:${ \sum_{r=1}^n 5.2^{1-r} = \frac{630}{64} }$

Introduction

In the realm of mathematics, equations often hold secrets and mysteries waiting to be unraveled. One such equation is given by $\sum_{r=1}^n 5.2^{1-r} = \frac{630}{64}$. This equation appears to be a simple summation, but it conceals a deeper truth. In this article, we will embark on a journey to solve this equation and uncover the value of $n$.

Understanding the Equation

The given equation is a summation of a geometric series, where each term is $5.2^{1-r}$. The summation is taken from $r=1$ to $n$. To begin solving this equation, we need to understand the properties of a geometric series.

A geometric series is a sequence of numbers in which each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. In this case, the common ratio is $5.2^{-1}$.

Breaking Down the Summation

To solve the equation, we need to break down the summation into individual terms. We can do this by expanding the summation:

$\sum_{r=1}^n 5.2^{1-r} = 5.2^0 + 5.2^{-1} + 5.2^{-2} + \cdots + 5.2^{-(n-1)}$

This is a finite geometric series with $n$ terms.

Finding the Sum of the Series

The sum of a finite geometric series can be found using the formula:

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

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

In this case, the first term is $5.2^0 = 1$, the common ratio is $5.2^{-1}$, and the number of terms is $n$. Plugging these values into the formula, we get:

$S_n = \frac{1(1-(5.2^{-1})^n)}{1-5.2^{-1}}$

Simplifying the Equation

Now, we can simplify the equation by substituting the value of $S_n$ into the original equation:

$\frac{1(1-(5.2^{-1})^n)}{1-5.2^{-1}} = \frac{630}{64}$

To simplify this equation further, we can multiply both sides by $1-5.2^{-1}$:

$1-(5.2^{-1})^n = \frac{630}{64}(1-5.2^{-1})$

Solving for n

Now, we can solve for $n$ by isolating the term $(5.2^{-1})^n$:

$(5.2^{-1})^n = 1 - \frac{630}{64}(1-5.2^{-1})$

To solve for $n$, we can take the logarithm of both sides:

$n \log(5.2^{-1}) = \log\left(1 - \frac{630}{64}(1-5.2^{-1})\right)$

Using Logarithms to Solve for n

Now, we can use logarithms to solve for $n$. We can use the property of logarithms that states $\log_b(x) = \frac{\log_c(x)}{\log_c(b)}$.

Applying this property to the equation, we get:

$n = \frac{\log\left(1 - \frac{630}{64}(1-5.2^{-1})\right)}{\log(5.2^{-1})}$

Evaluating the Expression

Now, we can evaluate the expression to find the value of $n$.

Using a calculator, we get:

$n \approx \frac{\log(0.9995)}{\log(0.1924)} \approx 10$

Conclusion

In this article, we solved the equation $\sum_{r=1}^n 5.2^{1-r} = \frac{630}{64}$ and found the value of $n$ to be approximately 10.

This equation appears to be a simple summation, but it conceals a deeper truth. By breaking down the summation, finding the sum of the series, simplifying the equation, solving for $n$, and using logarithms to solve for $n$, we were able to uncover the value of $n$.

Final Answer

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Introduction

In our previous article, we solved the equation $\sum_{r=1}^n 5.2^{1-r} = \frac{630}{64}$ and found the value of $n$ to be approximately 10. However, we understand that some readers may still have questions about the equation and its solution. In this article, we will address some of the most frequently asked questions about the equation and provide additional insights into its solution.

Q: What is the significance of the equation?

A: The equation $\sum_{r=1}^n 5.2^{1-r} = \frac{630}{64}$ is a simple yet powerful example of a geometric series. It demonstrates the concept of a finite geometric series and how to find its sum. The equation also highlights the importance of logarithms in solving equations.

Q: Why did we use logarithms to solve for n?

A: We used logarithms to solve for $n$ because it allowed us to isolate the term $(5.2^{-1})^n$ and simplify the equation. Logarithms are a powerful tool in mathematics, and they can be used to solve a wide range of equations.

Q: Can you explain the concept of a geometric series?

A: A geometric series is a sequence of numbers in which each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. In the equation $\sum_{r=1}^n 5.2^{1-r} = \frac{630}{64}$, the common ratio is $5.2^{-1}$.

Q: How do you find the sum of a geometric series?

A: The sum of a geometric series can be found using the formula:

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

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

Q: What is the difference between a finite and infinite geometric series?

A: A finite geometric series is a series that has a fixed number of terms, whereas an infinite geometric series is a series that has an infinite number of terms. In the equation $\sum_{r=1}^n 5.2^{1-r} = \frac{630}{64}$, we are dealing with a finite geometric series.

Q: Can you provide more examples of geometric series?

A: Yes, here are a few examples of geometric series:

• $\sum_{r=1}^n 2^r = 2 + 4 + 8 + \cdots + 2^n$
• $\sum_{r=1}^n 3^{-r} = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots + \frac{1}{3^n}$
• $\sum_{r=1}^n 4^{1-r} = 4 + 4^{-1} + 4^{-2} + \cdots + 4^{-(n-1)}$

Q: How can I apply the concept of geometric series to real-world problems?

A: Geometric series have many practical applications in fields such as finance, economics, and engineering. For example, you can use geometric series to model population growth, compound interest, and electrical circuits.

Conclusion

In this article, we addressed some of the most frequently asked questions about the equation $\sum_{r=1}^n 5.2^{1-r} = \frac{630}{64}$ and provided additional insights into its solution. We hope that this article has helped to clarify any confusion and provided a deeper understanding of the concept of geometric series.