Evaluate:$\[ \sum_{n=1}^{10} 8\left(\frac{1}{4}\right)^{n-1} \\]$\[ S = [?] \\]

by ADMIN 83 views

Introduction

In mathematics, a series is a sequence of numbers that are added together. A finite series is a series that has a finite number of terms, and it can be evaluated using various methods. In this article, we will evaluate the finite series ∑n=1108(14)n−1\sum_{n=1}^{10} 8\left(\frac{1}{4}\right)^{n-1} and find its sum.

Understanding the Series

The given series is a geometric series, which is a series of the form ∑n=1Narn−1\sum_{n=1}^{N} ar^{n-1}, where aa is the first term and rr is the common ratio. In this case, the first term aa is 88 and the common ratio rr is 14\frac{1}{4}.

Formula for the Sum of a Geometric Series

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

S=a(1−rN)1−rS = \frac{a(1-r^N)}{1-r}

where SS is the sum, aa is the first term, rr is the common ratio, and NN is the number of terms.

Evaluating the Series

We can now use the formula to evaluate the given series. Plugging in the values, we get:

S=8(1−(14)10)1−14S = \frac{8\left(1-\left(\frac{1}{4}\right)^{10}\right)}{1-\frac{1}{4}}

Simplifying the Expression

To simplify the expression, we can start by evaluating the exponent:

(14)10=11048576\left(\frac{1}{4}\right)^{10} = \frac{1}{1048576}

Now, we can plug this value back into the expression:

S=8(1−11048576)34S = \frac{8\left(1-\frac{1}{1048576}\right)}{\frac{3}{4}}

Further Simplification

To further simplify the expression, we can multiply the numerator and denominator by 44:

S=32(1−11048576)3S = \frac{32\left(1-\frac{1}{1048576}\right)}{3}

Evaluating the Expression

Now, we can evaluate the expression:

S=32(10485751048576)3S = \frac{32\left(\frac{1048575}{1048576}\right)}{3}

Simplifying the Fraction

To simplify the fraction, we can multiply the numerator and denominator by 10485761048576:

S=32(10485751048576)3=32(10485751048576)3S = \frac{32\left(\frac{1048575}{1048576}\right)}{3} = \frac{32\left(\frac{1048575}{1048576}\right)}{3}

Final Evaluation

To get the final answer, we can multiply the numerator and denominator by 33:

S=32(10485751048576)3=32(10485751048576)3S = \frac{32\left(\frac{1048575}{1048576}\right)}{3} = \frac{32\left(\frac{1048575}{1048576}\right)}{3}

Conclusion

In this article, we evaluated the finite series ∑n=1108(14)n−1\sum_{n=1}^{10} 8\left(\frac{1}{4}\right)^{n-1} and found its sum. The sum of the series is 32(10485751048576)3\frac{32\left(\frac{1048575}{1048576}\right)}{3}. This result can be used in various mathematical applications, such as in the calculation of interest rates and in the analysis of financial data.

Discussion

The evaluation of the series ∑n=1108(14)n−1\sum_{n=1}^{10} 8\left(\frac{1}{4}\right)^{n-1} is an important problem in mathematics, as it can be used to model real-world situations. For example, the series can be used to calculate the amount of money that will be accumulated in a savings account after a certain number of years, assuming a fixed interest rate.

Applications

The series ∑n=1108(14)n−1\sum_{n=1}^{10} 8\left(\frac{1}{4}\right)^{n-1} has many applications in mathematics and finance. Some of the applications include:

  • Interest Rates: The series can be used to calculate the interest rate on a savings account or a loan.
  • Financial Analysis: The series can be used to analyze financial data and make predictions about future trends.
  • Probability: The series can be used to calculate probabilities in probability theory.

Future Work

In the future, we can use the series ∑n=1108(14)n−1\sum_{n=1}^{10} 8\left(\frac{1}{4}\right)^{n-1} to model more complex real-world situations. For example, we can use the series to calculate the amount of money that will be accumulated in a savings account after a certain number of years, assuming a variable interest rate.

References

  • Mathematics: The series ∑n=1108(14)n−1\sum_{n=1}^{10} 8\left(\frac{1}{4}\right)^{n-1} is a classic problem in mathematics, and it has been studied by many mathematicians throughout history.
  • Finance: The series can be used to calculate interest rates and analyze financial data.

Conclusion

In conclusion, the evaluation of the series ∑n=1108(14)n−1\sum_{n=1}^{10} 8\left(\frac{1}{4}\right)^{n-1} is an important problem in mathematics, and it has many applications in finance and probability theory. The sum of the series is 32(10485751048576)3\frac{32\left(\frac{1048575}{1048576}\right)}{3}, and it can be used to model real-world situations and make predictions about future trends.

Introduction

In our previous article, we evaluated the finite series ∑n=1108(14)n−1\sum_{n=1}^{10} 8\left(\frac{1}{4}\right)^{n-1} and found its sum. In this article, we will answer some frequently asked questions about the series and its evaluation.

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

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

S=a(1−rN)1−rS = \frac{a(1-r^N)}{1-r}

where SS is the sum, aa is the first term, rr is the common ratio, and NN is the number of terms.

Q: How do I evaluate the series ∑n=1108(14)n−1\sum_{n=1}^{10} 8\left(\frac{1}{4}\right)^{n-1}?

A: To evaluate the series, you can use the formula for the sum of a geometric series. Plug in the values of aa, rr, and NN into the formula, and simplify the expression.

Q: What is the sum of the series ∑n=1108(14)n−1\sum_{n=1}^{10} 8\left(\frac{1}{4}\right)^{n-1}?

A: The sum of the series is 32(10485751048576)3\frac{32\left(\frac{1048575}{1048576}\right)}{3}.

Q: Can I use the series ∑n=1108(14)n−1\sum_{n=1}^{10} 8\left(\frac{1}{4}\right)^{n-1} to model real-world situations?

A: Yes, the series can be used to model real-world situations, such as calculating interest rates and analyzing financial data.

Q: What are some applications of the series ∑n=1108(14)n−1\sum_{n=1}^{10} 8\left(\frac{1}{4}\right)^{n-1}?

A: Some applications of the series include:

  • Interest Rates: The series can be used to calculate the interest rate on a savings account or a loan.
  • Financial Analysis: The series can be used to analyze financial data and make predictions about future trends.
  • Probability: The series can be used to calculate probabilities in probability theory.

Q: Can I use the series ∑n=1108(14)n−1\sum_{n=1}^{10} 8\left(\frac{1}{4}\right)^{n-1} to calculate the amount of money that will be accumulated in a savings account after a certain number of years?

A: Yes, the series can be used to calculate the amount of money that will be accumulated in a savings account after a certain number of years, assuming a fixed interest rate.

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

A: A geometric series is a series of the form ∑n=1Narn−1\sum_{n=1}^{N} ar^{n-1}, where aa is the first term and rr is the common ratio. An arithmetic series is a series of the form ∑n=1Na+(n−1)d\sum_{n=1}^{N} a + (n-1)d, where aa is the first term and dd is the common difference.

Q: Can I use the series ∑n=1108(14)n−1\sum_{n=1}^{10} 8\left(\frac{1}{4}\right)^{n-1} to calculate the amount of money that will be accumulated in a savings account after a certain number of years, assuming a variable interest rate?

A: No, the series is only valid for a fixed interest rate. To calculate the amount of money that will be accumulated in a savings account after a certain number of years, assuming a variable interest rate, you would need to use a different formula.

Conclusion

In this article, we answered some frequently asked questions about the series ∑n=1108(14)n−1\sum_{n=1}^{10} 8\left(\frac{1}{4}\right)^{n-1} and its evaluation. We hope that this article has been helpful in understanding the series and its applications.

Discussion

The series ∑n=1108(14)n−1\sum_{n=1}^{10} 8\left(\frac{1}{4}\right)^{n-1} is a classic problem in mathematics, and it has many applications in finance and probability theory. The sum of the series is 32(10485751048576)3\frac{32\left(\frac{1048575}{1048576}\right)}{3}, and it can be used to model real-world situations and make predictions about future trends.

References

  • Mathematics: The series ∑n=1108(14)n−1\sum_{n=1}^{10} 8\left(\frac{1}{4}\right)^{n-1} is a classic problem in mathematics, and it has been studied by many mathematicians throughout history.
  • Finance: The series can be used to calculate interest rates and analyze financial data.

Future Work

In the future, we can use the series ∑n=1108(14)n−1\sum_{n=1}^{10} 8\left(\frac{1}{4}\right)^{n-1} to model more complex real-world situations. For example, we can use the series to calculate the amount of money that will be accumulated in a savings account after a certain number of years, assuming a variable interest rate.

Conclusion

In conclusion, the series ∑n=1108(14)n−1\sum_{n=1}^{10} 8\left(\frac{1}{4}\right)^{n-1} is a classic problem in mathematics, and it has many applications in finance and probability theory. The sum of the series is 32(10485751048576)3\frac{32\left(\frac{1048575}{1048576}\right)}{3}, and it can be used to model real-world situations and make predictions about future trends.