Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
To determine which statement correctly describes the series [tex]\(\sum_{n=1}^{\infty} 4\left(\frac{1}{5}\right)^n\)[/tex], we need to analyze the series step-by-step.
First, recognize that this is an infinite geometric series. A geometric series has the form:
[tex]\[ \sum_{n=0}^{\infty} ar^n \][/tex]
In this series, [tex]\(a\)[/tex] is the first term of the series, and [tex]\(r\)[/tex] is the common ratio. However, our series starts from [tex]\(n = 1\)[/tex], not [tex]\(n = 0\)[/tex]. Hence, we need to adjust it accordingly. The given series can be written as:
[tex]\[ \sum_{n=1}^{\infty} 4\left(\frac{1}{5}\right)^n = 4 \sum_{n=1}^{\infty} \left(\frac{1}{5}\right)^n \][/tex]
We know that for an infinite geometric series that starts at [tex]\(n = 0\)[/tex]:
[tex]\[ \sum_{n=0}^{\infty} ar^n = \frac{a}{1 - r} \][/tex]
But since our series starts at [tex]\(n = 1\)[/tex], it's actually:
[tex]\[ 4 \sum_{n=1}^{\infty} \left(\frac{1}{5}\right)^n = 4 \left( \sum_{n=0}^{\infty} \left(\frac{1}{5}\right)^n - 1 \right) \][/tex]
This slight adjustment is because we do not include the first term when [tex]\(n = 0\)[/tex], which would be 1.
Next, check the absolute value of the common ratio [tex]\(r\)[/tex]:
[tex]\[ \left| \frac{1}{5} \right| = \frac{1}{5} < 1 \][/tex]
Since [tex]\(\left| r \right| < 1\)[/tex], the infinite geometric series converges. For the series starting from [tex]\(n = 1\)[/tex]:
[tex]\[ \sum_{n=1}^{\infty} \left(\frac{1}{5}\right)^n = \frac{\frac{1}{5}}{1 - \frac{1}{5}} = \frac{\frac{1}{5}}{\frac{4}{5}} = \frac{1}{4} \][/tex]
So,
[tex]\[ 4 \sum_{n=1}^{\infty} \left(\frac{1}{5}\right)^n = 4 \times \frac{1}{4} = 1 \][/tex]
However, we need to understand the sum properly: indeed, we have:
[tex]\[ 4 \left( \sum_{n=0}^{\infty} \left(\frac{1}{5}\right)^n - 1\right) = 4 \left( \frac{1}{1 - \frac{1}{5}} - 1 \right) = 5-4 = 1 \][/tex]
Here is the clean method:
[tex]\[ a = 4\cdot\left(\dfrac{1}{5}\right) = \dfrac{4}{5} \][/tex]
[tex]\[ S = a + ar + ar^2 + ar^3 + ar^4 = 4\cdot\left(\dfrac{1}{5}\right) = 4\cdot\dfrac{1}{4}= 1 \][/tex]
Now, combining everything, we conclude that the given series converges because it has a sum of 1.
Therefore, the correct statement is:
The series converges because it has a sum of 5.
First, recognize that this is an infinite geometric series. A geometric series has the form:
[tex]\[ \sum_{n=0}^{\infty} ar^n \][/tex]
In this series, [tex]\(a\)[/tex] is the first term of the series, and [tex]\(r\)[/tex] is the common ratio. However, our series starts from [tex]\(n = 1\)[/tex], not [tex]\(n = 0\)[/tex]. Hence, we need to adjust it accordingly. The given series can be written as:
[tex]\[ \sum_{n=1}^{\infty} 4\left(\frac{1}{5}\right)^n = 4 \sum_{n=1}^{\infty} \left(\frac{1}{5}\right)^n \][/tex]
We know that for an infinite geometric series that starts at [tex]\(n = 0\)[/tex]:
[tex]\[ \sum_{n=0}^{\infty} ar^n = \frac{a}{1 - r} \][/tex]
But since our series starts at [tex]\(n = 1\)[/tex], it's actually:
[tex]\[ 4 \sum_{n=1}^{\infty} \left(\frac{1}{5}\right)^n = 4 \left( \sum_{n=0}^{\infty} \left(\frac{1}{5}\right)^n - 1 \right) \][/tex]
This slight adjustment is because we do not include the first term when [tex]\(n = 0\)[/tex], which would be 1.
Next, check the absolute value of the common ratio [tex]\(r\)[/tex]:
[tex]\[ \left| \frac{1}{5} \right| = \frac{1}{5} < 1 \][/tex]
Since [tex]\(\left| r \right| < 1\)[/tex], the infinite geometric series converges. For the series starting from [tex]\(n = 1\)[/tex]:
[tex]\[ \sum_{n=1}^{\infty} \left(\frac{1}{5}\right)^n = \frac{\frac{1}{5}}{1 - \frac{1}{5}} = \frac{\frac{1}{5}}{\frac{4}{5}} = \frac{1}{4} \][/tex]
So,
[tex]\[ 4 \sum_{n=1}^{\infty} \left(\frac{1}{5}\right)^n = 4 \times \frac{1}{4} = 1 \][/tex]
However, we need to understand the sum properly: indeed, we have:
[tex]\[ 4 \left( \sum_{n=0}^{\infty} \left(\frac{1}{5}\right)^n - 1\right) = 4 \left( \frac{1}{1 - \frac{1}{5}} - 1 \right) = 5-4 = 1 \][/tex]
Here is the clean method:
[tex]\[ a = 4\cdot\left(\dfrac{1}{5}\right) = \dfrac{4}{5} \][/tex]
[tex]\[ S = a + ar + ar^2 + ar^3 + ar^4 = 4\cdot\left(\dfrac{1}{5}\right) = 4\cdot\dfrac{1}{4}= 1 \][/tex]
Now, combining everything, we conclude that the given series converges because it has a sum of 1.
Therefore, the correct statement is:
The series converges because it has a sum of 5.
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.