(f) $$ \sum_{\mathrm{r}=0}^{\mathrm{x}} \mathrm{a}^{\mathrm{r}}(-1)^{\mathrm{r}-1} $$;

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To evaluate the finite sum

$$
\sum_{r=0}^{x} a^{r} (-1)^{r - 1},
$$

we can simplify it using the properties of geometric series. Here's the step-by-step process:

1. **Rewrite the Sum:**

$$
   \sum_{r=0}^{x} a^{r} (-1)^{r - 1} = (-1)^{-1} \sum_{r=0}^{x} (-a)^{r} = -\sum_{r=0}^{x} (-a)^{r}
   $$

2. **Recognize the Geometric Series:**

The sum $$\sum_{r=0}^{x} (-a)^{r}$$ is a geometric series with common ratio $$q = -a$$. The formula for the sum of the first $$x + 1$$ terms of a geometric series is:

$$
   \sum_{r=0}^{x} q^{r} = \frac{1 - q^{x + 1}}{1 - q}, \quad \text{for } q \neq 1.
   $$

3. **Apply the Geometric Series Formula:**

Substituting $$q = -a$$:

$$
   \sum_{r=0}^{x} (-a)^{r} = \frac{1 - (-a)^{x + 1}}{1 - (-a)} = \frac{1 - (-a)^{x + 1}}{1 + a}.
   $$

4. **Include the Earlier Factor:**

Recall that:

$$
   \sum_{r=0}^{x} a^{r} (-1)^{r - 1} = -\sum_{r=0}^{x} (-a)^{r} = -\left( \frac{1 - (-a)^{x + 1}}{1 + a} \right).
   $$

5. **Simplify the Expression:**

$$
   -\left( \frac{1 - (-a)^{x + 1}}{1 + a} \right) = \frac{(-a)^{x + 1} - 1}{1 + a}.
   $$

**Final Result:**

$$
\sum_{r=0}^{x} a^{r} (-1)^{r - 1} = \frac{(-a)^{x + 1} - 1}{1 + a}.
$$

This closed-form expression provides a simplified way to compute the sum for any integer $$x \geq 0$$ and $$a \neq -1$$.
The sum simplifies to $$\frac{(-a)^{x + 1} - 1}{1 + a}$$.

(f) ;

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