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Homework:HW SECTION 10.1
Question 6, 10.1.67
Part 1 of 4
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Part 1
Consider the following infinite series. Complete parts​ (a) through​ (c) below.
Summation from k equals 1 to infinity StartFraction 64 Over left parenthesis 2 k minus 1 right parenthesis left parenthesis 2 k plus 1 right parenthesis EndFraction
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Part 1
a. Find the first four partial sums Upper S 1​, Upper S 2​, Upper S 3​, and Upper S 4 of the series.
Upper S 1equals
  
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Upper S 2equals
  
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Upper S 3equals
  
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Upper S 4equals
  
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​(Type integers or simplified​ fractions.)
Summation from k equals 1 to infinity StartFraction 64 Over left parenthesis 2 k minus 1 right parenthesis left parenthesis 2 k plus 1 right parenthesis EndFractionSummation from k equals 1 to infinity StartFraction 64 Over left parenthesis 2 k minus 1 right parenthesis left parenthesis 2 k plus 1 right parenthesis EndFractionSummation from k equals 1 to infinity StartFraction 64 Over left parenthesis 2 k minus 1 right parenthesis left parenthesis 2 k plus 1 right parenthesis EndFractionSummation from k equals 1 to infinity StartFraction 64 Over left parenthesis 2 k minus 1 right parenthesis left parenthesis 2 k plus 1 right parenthesis EndFraction
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Let's break down the problem and find the first four partial sums \( S_1 \), \( S_2 \), \( S_3 \), and \( S_4 \) for the infinite series.

The series given is:

\[
\sum_{k=1}^{\infty} \frac{64}{(2k-1)(2k+1)}
\]

### Step 1: Simplify the General Term

First, we need to simplify the general term:

\[
\frac{64}{(2k-1)(2k+1)}
\]

Notice that this can be decomposed into partial fractions:

\[
\frac{64}{(2k-1)(2k+1)} = \frac{A}{2k-1} + \frac{B}{2k+1}
\]

To find \( A \) and \( B \), multiply both sides by \( (2k-1)(2k+1) \):

\[
64 = A(2k+1) + B(2k-1)
\]

Expanding and equating coefficients of like terms:

\[
64 = (2A + 2B)k + (A - B)
\]

From this equation, we get two separate equations:

\[
2A + 2B = 0 \quad \text{(1)}
\]
\[
A - B = 64 \quad \text{(2)}
\]

From equation (1), \( A + B = 0 \), so \( B = -A \).

Substituting \( B = -A \) into equation (2):

\[
A - (-A) = 64 \implies 2A = 64 \implies A = 32
\]

So, \( B = -32 \).

Thus, the partial fraction decomposition is:

\[
\frac{64}{(2k-1)(2k+1)} = \frac{32}{2k-1} - \frac{32}{2k+1}
\]

### Step 2: Write out the First Few Terms

Now, we can write the series as:

\[
\sum_{k=1}^{\infty} \left(\frac{32}{2k-1} - \frac{32}{2k+1}\right)
\]

The first few terms of the series are:

- For \( k = 1 \):
  \[
  \frac{32}{1} - \frac{32}{3} = 32 - \frac{32}{3} = \frac{96}{3} - \frac{32}{3} = \frac{64}{3}
  \]

- For \( k = 2 \):
  \[
  \frac{32}{3} - \frac{32}{5} = \frac{160}{15} - \frac{96}{15} = \frac{64}{15}
  \]

- For \( k = 3 \):
  \[
  \frac{32}{5} - \frac{32}{7} = \frac{224}{35} - \frac{160}{35} = \frac{64}{35}
  \]

- For \( k = 4 \):
  \[
  \frac{32}{7} - \frac{32}{9} = \frac{288}{63} - \frac{224}{63} = \frac{64}{63}
  \]

### Step 3: Calculate the Partial Sums

- \( S_1 = \frac{64}{3} \)

- \( S_2 = \frac{64}{3} + \frac{64}{15} = \frac{320}{15} + \frac{64}{15} = \frac{384}{15} = \frac{128}{5} \)

- \( S_3 = \frac{128}{5} + \frac{64}{35} = \frac{896}{35} + \frac{64}{35} = \frac{960}{35} = \frac{192}{7} \)

- \( S_4 = \frac{192}{7} + \frac{64}{63} = \frac{12096}{441} + \frac{448}{441} = \frac{12544}{441} \)

So, the first four partial sums are:

\[
\begin{aligned}
S_1 & = \frac{64}{3}, \\
S_2 & = \frac{128}{5}, \\
S_3 & = \frac{192}{7}, \\
S_4 & = \frac{12544}{441}.
\end{aligned}
\]

Would you like to explore this further or have any questions? Here are some related questions you might consider:

1. How would the partial sums change if the initial series was different?
2. What is the general pattern for the partial sums in telescoping series?
3. Can you derive a general formula for \( S_n \) for this series?
4. What would happen if the series started at a different index, say \( k = 2 \)?
5. How does the convergence of this series compare to other series like geometric or harmonic series?

**Tip:** In telescoping series, it's often useful to simplify the series to see how terms cancel each other out, making it easier to find partial sums.

Consider the following infinite series. Complete parts​ (a) through​ (c) below. Summation from k equals 1 to infinity StartFraction 64 Over left parenthesis 2 k minus 1 right parenthesis left parenthesis 2 k plus 1 right parenthesis EndFraction. a. Find the first four partial sums Upper S 1​, Upper S 2​, Upper S 3​, and Upper S 4 of the series.

Explore how to find the first four partial sums for the telescoping series involving the summation of 64 over the product of (2k-1) and (2k+1). This problem includes key concepts from calculus, such as partial sums and partial fraction decomposition. Suitable for undergraduate students.

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