Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
b. Generalize the Results in the Table
To generalize the results, we start with understanding the geometric series at stage [tex]\( n \)[/tex].
At stage [tex]\( n \)[/tex], the second column gives us the terms of the geometric series:
[tex]\[ \frac{3}{4} + \frac{3}{16} + \cdots + 3 \left(\frac{1}{4}\right)^n \][/tex]
This series can be expressed as:
[tex]\[ a + ar + ar^2 + \cdots + ar^n \][/tex]
Where:
- [tex]\( a = \frac{3}{4} \)[/tex] (the first term)
- [tex]\( r = \frac{1}{4} \)[/tex] (the common ratio)
The sum of the first [tex]\( n \)[/tex] terms of a finite geometric series is given by the formula:
[tex]\[ S_n = a \frac{1 - r^{n+1}}{1 - r} \][/tex]
By substituting [tex]\( a = \frac{3}{4} \)[/tex] and [tex]\( r = \frac{1}{4} \)[/tex] into the formula, we get the specific formula that the third column gives you:
[tex]\[ S_n = \frac{3}{4} \frac{1 - \left(\frac{1}{4}\right)^{n+1}}{1 - \frac{1}{4}} \][/tex]
Simplifying this, we obtain:
[tex]\[ S_n = \frac{3}{4} \frac{1 - \left(\frac{1}{4}\right)^{n+1}}{\frac{3}{4}} \][/tex]
This simplifies further to:
[tex]\[ S_n = 3 \left(1 - \left(\frac{1}{4}\right)^{n+1}\right) \][/tex]
So the formula given in the third column is:
[tex]\[ S_n = 3 \left(1 - \left(\frac{1}{4}\right)^{n+1}\right) \][/tex]
c. Show that the General Formula for the Sum of a Finite Geometric Series Agrees with the Specific Formula
To show that the general formula for the sum of a finite geometric series [tex]\( S_n = a \frac{1 - r^{n+1}}{1 - r} \)[/tex] agrees with the specific formula [tex]\( S_n = 3 \left(1 - (\frac{1}{4})^{n+1}\right) \)[/tex], we need to compare these formulas for specific terms.
For the general formula:
[tex]\[ S_n = a \frac{1 - r^{n+1}}{1 - r} \][/tex]
Where [tex]\( a = \frac{3}{4} \)[/tex] and [tex]\( r = \frac{1}{4} \)[/tex], we substitute these values:
[tex]\[ S_n = \frac{3}{4} \frac{1 - (\frac{1}{4})^{n+1}}{1 - \frac{1}{4}} \][/tex]
[tex]\[ S_n = \frac{3}{4} \frac{1 - (\frac{1}{4})^{n+1}}{\frac{3}{4}} \][/tex]
[tex]\[ S_n = 3 \left(1 - (\frac{1}{4})^{n+1}\right) \][/tex]
Therefore, the general formula
[tex]\[ S_n = a \frac{1 - r^{n+1}}{1 - r} \][/tex]
for [tex]\( a = \frac{3}{4} \)[/tex] and [tex]\( r = \frac{1}{4} \)[/tex] is indeed equal to the specific formula:
[tex]\[ S_n = 3(1 - (\frac{1}{4})^{n+1}) \][/tex]
Hence, the general formula for the sum of a finite geometric series perfectly agrees with the specific formula derived from part (b).
To generalize the results, we start with understanding the geometric series at stage [tex]\( n \)[/tex].
At stage [tex]\( n \)[/tex], the second column gives us the terms of the geometric series:
[tex]\[ \frac{3}{4} + \frac{3}{16} + \cdots + 3 \left(\frac{1}{4}\right)^n \][/tex]
This series can be expressed as:
[tex]\[ a + ar + ar^2 + \cdots + ar^n \][/tex]
Where:
- [tex]\( a = \frac{3}{4} \)[/tex] (the first term)
- [tex]\( r = \frac{1}{4} \)[/tex] (the common ratio)
The sum of the first [tex]\( n \)[/tex] terms of a finite geometric series is given by the formula:
[tex]\[ S_n = a \frac{1 - r^{n+1}}{1 - r} \][/tex]
By substituting [tex]\( a = \frac{3}{4} \)[/tex] and [tex]\( r = \frac{1}{4} \)[/tex] into the formula, we get the specific formula that the third column gives you:
[tex]\[ S_n = \frac{3}{4} \frac{1 - \left(\frac{1}{4}\right)^{n+1}}{1 - \frac{1}{4}} \][/tex]
Simplifying this, we obtain:
[tex]\[ S_n = \frac{3}{4} \frac{1 - \left(\frac{1}{4}\right)^{n+1}}{\frac{3}{4}} \][/tex]
This simplifies further to:
[tex]\[ S_n = 3 \left(1 - \left(\frac{1}{4}\right)^{n+1}\right) \][/tex]
So the formula given in the third column is:
[tex]\[ S_n = 3 \left(1 - \left(\frac{1}{4}\right)^{n+1}\right) \][/tex]
c. Show that the General Formula for the Sum of a Finite Geometric Series Agrees with the Specific Formula
To show that the general formula for the sum of a finite geometric series [tex]\( S_n = a \frac{1 - r^{n+1}}{1 - r} \)[/tex] agrees with the specific formula [tex]\( S_n = 3 \left(1 - (\frac{1}{4})^{n+1}\right) \)[/tex], we need to compare these formulas for specific terms.
For the general formula:
[tex]\[ S_n = a \frac{1 - r^{n+1}}{1 - r} \][/tex]
Where [tex]\( a = \frac{3}{4} \)[/tex] and [tex]\( r = \frac{1}{4} \)[/tex], we substitute these values:
[tex]\[ S_n = \frac{3}{4} \frac{1 - (\frac{1}{4})^{n+1}}{1 - \frac{1}{4}} \][/tex]
[tex]\[ S_n = \frac{3}{4} \frac{1 - (\frac{1}{4})^{n+1}}{\frac{3}{4}} \][/tex]
[tex]\[ S_n = 3 \left(1 - (\frac{1}{4})^{n+1}\right) \][/tex]
Therefore, the general formula
[tex]\[ S_n = a \frac{1 - r^{n+1}}{1 - r} \][/tex]
for [tex]\( a = \frac{3}{4} \)[/tex] and [tex]\( r = \frac{1}{4} \)[/tex] is indeed equal to the specific formula:
[tex]\[ S_n = 3(1 - (\frac{1}{4})^{n+1}) \][/tex]
Hence, the general formula for the sum of a finite geometric series perfectly agrees with the specific formula derived from part (b).
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.