IDNLearn.com: Your go-to resource for finding precise and accurate answers. Ask your questions and receive reliable and comprehensive answers from our dedicated community of professionals.
Sagot :
Sure, let's find the sum \( S_n \) for each geometric series case described.
The formula for the sum of the first \( n \) terms of a geometric series is:
[tex]\[ S_n = \begin{cases} a_1 \cdot \dfrac{1 - r^n}{1 - r} & \text{if } r \neq 1 \\ a_1 \cdot n & \text{if } r = 1 \end{cases} \][/tex]
### Case 1: \( a_1 = 2, a_n = 486, r = 3 \)
1. First, we need to find \( n \) (the number of terms):
[tex]\[ a_n = a_1 \cdot r^{n-1} \\ 486 = 2 \cdot 3^{n-1} \\ 243 = 3^{n-1} \][/tex]
Since \( 243 = 3^5 \), then \( n-1 = 5 \) and \( n = 6 \).
2. Now, calculate the sum \( S_6 \):
[tex]\[ S_6 = 2 \cdot \left(\frac{1 - 3^6}{1 - 3}\right) \\ = 2 \cdot \left(\frac{1 - 729}{1 - 3}\right) \\ = 2 \cdot \left(\frac{-728}{-2}\right) \\ = 2 \cdot 364 \\ = 728 \][/tex]
### Case 2: \( a_1 = 1200, a_n = 75, r = \frac{1}{2} \)
1. First, we need to find \( n \):
[tex]\[ a_n = a_1 \cdot r^{n-1} \\ 75 = 1200 \cdot \left(\frac{1}{2}\right)^{n-1} \\ \left(\frac{1}{2}\right)^{n-1} = \frac{75}{1200} \\ \left(\frac{1}{2}\right)^{n-1} = \frac{1}{16} \\ \left(\frac{1}{2}\right)^{n-1} = \left(\frac{1}{2}\right)^4 \][/tex]
Here, \( n-1 = 4 \) and \( n = 5 \).
2. Now, calculate the sum \( S_5 \):
[tex]\[ S_5 = 1200 \cdot \left(\frac{1 - \left(\frac{1}{2}\right)^5}{1 - \frac{1}{2}}\right) \\ = 1200 \cdot \left(\frac{1 - \frac{1}{32}}{\frac{1}{2}}\right) \\ = 1200 \cdot \left(\frac{\frac{31}{32}}{\frac{1}{2}}\right) \\ = 1200 \cdot \left(\frac{31}{16}\right) \\ = 1200 \cdot 1.9375 \\ = 2325 \][/tex]
### Case 4: \( a_1 = 3, r = \frac{1}{3}, n = 4 \)
1. Using the sum formula for \( n \) terms:
[tex]\[ S_4 = 3 \cdot \left(\frac{1 - \left(\frac {1}{3}\right)^4}{1 - \frac{1}{3}}\right) \\ = 3 \cdot \left(\frac{1 - \frac{1}{81}}{\frac{2}{3}}\right) \\ = 3 \cdot \left(\frac{\frac{80}{81}}{\frac{2}{3}}\right) \\ = 3 \cdot \left(\frac{80}{81} \cdot \frac{3}{2}\right) \\ = 3 \cdot \frac {240}{162} \\ = 3 \cdot \frac{120}{81} \\ = 4.444 \approx 4.44 \][/tex]
### Case 5: \( a_1 = 2, r = 6, n = 4 \)
1. Using the sum formula for \( n \) terms:
[tex]\[ S_4 = 2 \cdot \left(\frac{1 - 6^4}{1 - 6}\right) \\ = 2 \cdot \left(\frac{1 - 1296}{-5}\right) \\ = 2 \cdot \left(\frac{-1295}{-5}\right) \\ = 2 \cdot 259 \\ = 518 \][/tex]
### Case 7: \( a_1 = 100, r = - \frac{1}{2}, n = 5 \)
1. Using the sum formula for \( n \) terms:
[tex]\[ S_5 = 100 \cdot \left(\frac{1 - \left(-\frac {1}{2}\right)^5}{1 - \left(-\frac {1}{2}\right)}\right) \\ = 100 \cdot \left(\frac{1 - \left(-\frac {1}{32}\right)}{1 + \frac{1}{2}}\right) \\ = 100 \cdot \left(\frac{1 - \left(-0.03125\right)}{1 + 0.5}\right) \\ = 100 \cdot \left(\frac{1 + 0.03125}{1.5}\right) \\ = 100 \cdot \left(\frac{1.03125}{1.5}\right) \\ \approx 68.75 \][/tex]
### Case 8: \( a_0 = 20 \) (initial term without ratio)
This case seems to refer a merely a single term geometric 0-th term series which has nothing further to sum just equal to 20
By keeping our interpretation \( a_{0}\ term \)
### Summarized Solutions:
1. \( S_1 = 728 \)
2. \( S_2 = 2325 \)
4. \( S_4 = 4.44 \)
5. \( S_5 = 518 \)
7. \( S_7 = 68.75 \)
8. \( S_{8}=20\
The formula for the sum of the first \( n \) terms of a geometric series is:
[tex]\[ S_n = \begin{cases} a_1 \cdot \dfrac{1 - r^n}{1 - r} & \text{if } r \neq 1 \\ a_1 \cdot n & \text{if } r = 1 \end{cases} \][/tex]
### Case 1: \( a_1 = 2, a_n = 486, r = 3 \)
1. First, we need to find \( n \) (the number of terms):
[tex]\[ a_n = a_1 \cdot r^{n-1} \\ 486 = 2 \cdot 3^{n-1} \\ 243 = 3^{n-1} \][/tex]
Since \( 243 = 3^5 \), then \( n-1 = 5 \) and \( n = 6 \).
2. Now, calculate the sum \( S_6 \):
[tex]\[ S_6 = 2 \cdot \left(\frac{1 - 3^6}{1 - 3}\right) \\ = 2 \cdot \left(\frac{1 - 729}{1 - 3}\right) \\ = 2 \cdot \left(\frac{-728}{-2}\right) \\ = 2 \cdot 364 \\ = 728 \][/tex]
### Case 2: \( a_1 = 1200, a_n = 75, r = \frac{1}{2} \)
1. First, we need to find \( n \):
[tex]\[ a_n = a_1 \cdot r^{n-1} \\ 75 = 1200 \cdot \left(\frac{1}{2}\right)^{n-1} \\ \left(\frac{1}{2}\right)^{n-1} = \frac{75}{1200} \\ \left(\frac{1}{2}\right)^{n-1} = \frac{1}{16} \\ \left(\frac{1}{2}\right)^{n-1} = \left(\frac{1}{2}\right)^4 \][/tex]
Here, \( n-1 = 4 \) and \( n = 5 \).
2. Now, calculate the sum \( S_5 \):
[tex]\[ S_5 = 1200 \cdot \left(\frac{1 - \left(\frac{1}{2}\right)^5}{1 - \frac{1}{2}}\right) \\ = 1200 \cdot \left(\frac{1 - \frac{1}{32}}{\frac{1}{2}}\right) \\ = 1200 \cdot \left(\frac{\frac{31}{32}}{\frac{1}{2}}\right) \\ = 1200 \cdot \left(\frac{31}{16}\right) \\ = 1200 \cdot 1.9375 \\ = 2325 \][/tex]
### Case 4: \( a_1 = 3, r = \frac{1}{3}, n = 4 \)
1. Using the sum formula for \( n \) terms:
[tex]\[ S_4 = 3 \cdot \left(\frac{1 - \left(\frac {1}{3}\right)^4}{1 - \frac{1}{3}}\right) \\ = 3 \cdot \left(\frac{1 - \frac{1}{81}}{\frac{2}{3}}\right) \\ = 3 \cdot \left(\frac{\frac{80}{81}}{\frac{2}{3}}\right) \\ = 3 \cdot \left(\frac{80}{81} \cdot \frac{3}{2}\right) \\ = 3 \cdot \frac {240}{162} \\ = 3 \cdot \frac{120}{81} \\ = 4.444 \approx 4.44 \][/tex]
### Case 5: \( a_1 = 2, r = 6, n = 4 \)
1. Using the sum formula for \( n \) terms:
[tex]\[ S_4 = 2 \cdot \left(\frac{1 - 6^4}{1 - 6}\right) \\ = 2 \cdot \left(\frac{1 - 1296}{-5}\right) \\ = 2 \cdot \left(\frac{-1295}{-5}\right) \\ = 2 \cdot 259 \\ = 518 \][/tex]
### Case 7: \( a_1 = 100, r = - \frac{1}{2}, n = 5 \)
1. Using the sum formula for \( n \) terms:
[tex]\[ S_5 = 100 \cdot \left(\frac{1 - \left(-\frac {1}{2}\right)^5}{1 - \left(-\frac {1}{2}\right)}\right) \\ = 100 \cdot \left(\frac{1 - \left(-\frac {1}{32}\right)}{1 + \frac{1}{2}}\right) \\ = 100 \cdot \left(\frac{1 - \left(-0.03125\right)}{1 + 0.5}\right) \\ = 100 \cdot \left(\frac{1 + 0.03125}{1.5}\right) \\ = 100 \cdot \left(\frac{1.03125}{1.5}\right) \\ \approx 68.75 \][/tex]
### Case 8: \( a_0 = 20 \) (initial term without ratio)
This case seems to refer a merely a single term geometric 0-th term series which has nothing further to sum just equal to 20
By keeping our interpretation \( a_{0}\ term \)
### Summarized Solutions:
1. \( S_1 = 728 \)
2. \( S_2 = 2325 \)
4. \( S_4 = 4.44 \)
5. \( S_5 = 518 \)
7. \( S_7 = 68.75 \)
8. \( S_{8}=20\
Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. Your questions find clarity at IDNLearn.com. Thanks for stopping by, and come back for more dependable solutions.
