
Homework:HW SECTION 10.6
Question 3, 10.6.30
Part 1 of 2
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Part 1
For the convergent alternating series Summation from k equals 0 to infinity StartFraction left parenthesis negative 1 right parenthesis Superscript k Over left parenthesis 4 k plus 2 right parenthesis Superscript 4 EndFraction
​, evaluate the nth partial sum for nequals2. Then find an upper bound for the error StartAbsoluteValue Upper S minus Upper S Subscript n EndAbsoluteValue in using the nth partial sum Upper S Subscript n to estimate the value of the series S.
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Part 1
The nth partial sum for the given value of n is 
  
enter your response here.
​(Type an integer or a decimal. Round to seven decimal places as​ needed.)
The nth partial sum for the given value of n is(Type an integer or a decimal. Round to seven decimal places as needed.)Using the theorem, the upper bound for the error is(Type an integer or a decimal. Round to seven decimal places as needed.)The nth partial sum for the given value of n is(Type an integer or a decimal. Round to seven decimal places as needed.)
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Let's solve the problem step by step.

### Problem Statement Recap
We are given the following alternating series:

\[
\sum_{k=0}^{\infty} \frac{(-1)^k}{(4k + 2)^4}
\]

We need to:

1. Find the 2nd partial sum \( S_2 \).
2. Determine an upper bound for the error \( |S - S_n| \) when approximating the series \( S \) with the \( n \)-th partial sum \( S_n \), for \( n = 2 \).

### Step 1: Finding the 2nd Partial Sum \( S_2 \)
The nth partial sum \( S_n \) for an alternating series is the sum of the first \( n \) terms of the series.

For \( n = 2 \), we calculate the partial sum \( S_2 \):

\[
S_2 = \frac{(-1)^0}{(4(0) + 2)^4} + \frac{(-1)^1}{(4(1) + 2)^4}
\]

This simplifies to:

\[
S_2 = \frac{1}{2^4} - \frac{1}{6^4}
\]

Let's calculate this:

\[
\frac{1}{2^4} = \frac{1}{16} = 0.0625
\]
\[
\frac{1}{6^4} = \frac{1}{1296} \approx 0.0007716
\]

So:

\[
S_2 = 0.0625 - 0.0007716 \approx 0.0617284
\]

### Step 2: Finding the Upper Bound for the Error
The error bound for an alternating series is given by the absolute value of the first omitted term:

\[
|S - S_2| \leq \left| \frac{(-1)^2}{(4(2) + 2)^4} \right|
\]

Simplifying the expression inside the absolute value:

\[
\left| \frac{1}{(8 + 2)^4} \right| = \frac{1}{10^4} = \frac{1}{10000} = 0.0001
\]

### Final Answers
- The 2nd partial sum \( S_2 \) is approximately **0.0617284**.
- The upper bound for the error \( |S - S_2| \) is **0.0001**.

Do you want details or have any questions?

Here are 5 related questions you might want to consider:
1. What happens to the partial sum as \( n \) increases?
2. How can the error bound be decreased further?
3. What is the general form of the \( n \)-th term of this series?
4. How does the error bound affect the accuracy of the partial sum?
5. Can this series be expressed in a closed form?

**Tip:** When evaluating alternating series, the error bound is a useful tool to determine how accurate your partial sum is.

For the convergent alternating series Summation from k equals 0 to infinity StartFraction left parenthesis negative 1 right parenthesis Superscript k Over left parenthesis 4 k plus 2 right parenthesis Superscript 4 EndFraction​, evaluate the nth partial sum for n equals 2. Then find an upper bound for the error StartAbsoluteValue Upper S minus Upper S Subscript n EndAbsoluteValue in using the nth partial sum Upper S Subscript n to estimate the value of the series S.

Explore the process of evaluating the nth partial sum and determining the upper bound for the error in a convergent alternating series. This detailed solution covers series and sequences, alternating series, partial sums, and error bounds, suitable for advanced high school students (Grades 11-12).

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