Math Problem Statement

Consider the finite geometric series. βˆ‘ π‘˜

1 𝑛 4 ( 1 3 ) π‘˜ βˆ’ 1 k=1 βˆ‘ n ​ 4( 3 1 ​ ) kβˆ’1

Evaluate the series for each of the given values of 𝑛 n​. You may round your answers to five decimal places.

𝑛

6 n=6.​ Preview ​ 𝑛

8 n=8​. Preview ​ 𝑛

10 n=10​.

Solution

To evaluate the given finite geometric series for different values of nn, we can use the formula for the sum of a geometric series:

Sn=a1βˆ’rn1βˆ’rS_n = a \frac{1 - r^n}{1 - r}

where:

  • aa is the first term of the series,
  • rr is the common ratio, and
  • nn is the number of terms.

For the series βˆ‘k=1n4(13)kβˆ’1\sum_{k=1}^{n} 4\left(\frac{1}{3}\right)^{k-1}, the first term aa is 4 (when k=1k = 1) and the common ratio rr is 13\frac{1}{3}.

Let's calculate the sum for each value of nn.

  1. For n=6n = 6:

    • a=4a = 4
    • r=13r = \frac{1}{3}
    • n=6n = 6

    S6=41βˆ’(13)61βˆ’13S_6 = 4 \frac{1 - \left(\frac{1}{3}\right)^6}{1 - \frac{1}{3}}

  2. For n=8n = 8:

    • a=4a = 4
    • r=13r = \frac{1}{3}
    • n=8n = 8

    S8=41βˆ’(13)81βˆ’13S_8 = 4 \frac{1 - \left(\frac{1}{3}\right)^8}{1 - \frac{1}{3}}

  3. For n=10n = 10:

    • a=4a = 4
    • r=13r = \frac{1}{3}
    • n=10n = 10

    S10=41βˆ’(13)101βˆ’13S_{10} = 4 \frac{1 - \left(\frac{1}{3}\right)^{10}}{1 - \frac{1}{3}}

Let's calculate these values.Here are the evaluated sums for the geometric series with different values of nn, rounded to five decimal places:

  1. For n=6n = 6: S6β‰ˆ5.99177S_6 \approx 5.99177
  2. For n=8n = 8: S8β‰ˆ5.99909S_8 \approx 5.99909
  3. For n=10n = 10: S10β‰ˆ5.99990S_{10} \approx 5.99990

Would you like any further details or explanations?

Here are some related questions to consider:

  1. How does the common ratio affect the convergence of a geometric series?
  2. What happens to the sum as nn approaches infinity for this series?
  3. How would the sum change if the first term aa were different?
  4. What is the significance of the common ratio being less than 1?
  5. Can a geometric series have a negative common ratio? If so, how does it affect the series?
  6. What if the common ratio was greater than 1? How would that affect the series?
  7. How can we use geometric series in real-world applications?
  8. What are the conditions for a geometric series to converge?

Tip: Understanding the behavior of a geometric series as the number of terms increases can help in approximating the sum for large nn.

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Math Problem Analysis

Mathematical Concepts

Geometric Series

Formulas

Sum of a geometric series: \( S_n = a \frac{1 - r^n}{1 - r} \)

Theorems

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Suitable Grade Level

Advanced High School