Let's solve the problem step by step. The series is given as:

$\sum_{n=1}^\infty \frac{10}{(-3)^n}.$

Part (a): First four terms of the sequence $\{a_n\}_{n=1}^\infty$.

The general term of the sequence is:

$a_n = \frac{10}{(-3)^n}.$

Calculate the first four terms:

1. $a_1 = \frac{10}{(-3)^1} = \frac{10}{-3} = -\frac{10}{3}$
2. $a_2 = \frac{10}{(-3)^2} = \frac{10}{9}$
3. $a_3 = \frac{10}{(-3)^3} = \frac{10}{-27} = -\frac{10}{27}$
4. $a_4 = \frac{10}{(-3)^4} = \frac{10}{81}$

So, the first four terms are: $a_1 = -\frac{10}{3}, \quad a_2 = \frac{10}{9}, \quad a_3 = -\frac{10}{27}, \quad a_4 = \frac{10}{81}.$

Part (b): First four terms of the sequence of partial sums $\{S_n\}_{n=1}^\infty$.

The partial sum $S_n$ is:

$S_n = a_1 + a_2 + \cdots + a_n.$

Compute the first four partial sums:

1. $S_1 = a_1 = -\frac{10}{3}$
2. $S_2 = a_1 + a_2 = -\frac{10}{3} + \frac{10}{9} = -\frac{30}{9} + \frac{10}{9} = -\frac{20}{9}$
3. $S_3 = S_2 + a_3 = -\frac{20}{9} + \left(-\frac{10}{27}\right) = -\frac{60}{27} - \frac{10}{27} = -\frac{70}{27}$
4. $S_4 = S_3 + a_4 = -\frac{70}{27} + \frac{10}{81} = -\frac{210}{81} + \frac{10}{81} = -\frac{200}{81}$

So, the partial sums are: $S_1 = -\frac{10}{3}, \quad S_2 = -\frac{20}{9}, \quad S_3 = -\frac{70}{27}, \quad S_4 = -\frac{200}{81}.$

Part (c): Recursive formula for $S_k$, the $k$-th partial sum.

The recursive relation for the partial sum can be derived as: $S_k = S_{k-1} + a_k.$

From the series definition: $a_k = \frac{10}{(-3)^k}.$

Thus, the recursive formula is: $S_k = S_{k-1} + \frac{10}{(-3)^k}.$

If you'd like, I can verify or simplify further. Would you like more details or explanations?

Five Related Questions:

1. What happens to the partial sums $S_n$ as $n \to \infty$?
2. Can this series be rewritten in a closed form using the sum of a geometric series?
3. How does the convergence of the series depend on the base $-3$?
4. What is the general formula for $S_n$ without recursion?
5. How would the series change if the numerator was not constant (e.g., $10n$)?

Tip:

For alternating series like this one, consider using the geometric series formula when possible—it simplifies calculations significantly!