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:

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.