2.1 Determine $$ n $$ if $$ \sum_{k=1}^{n}(-8)\left(\frac{1}{2}\right)^{k-1}=-\frac{63}{4} $$

Question

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To determine the value of $$ n $$ in the equation:

$$
\sum_{k=1}^{n} (-8) \left( \frac{1}{2} \right)^{k-1} = -\frac{63}{4}
$$

we recognize that this is the sum of a geometric series. The general formula for the sum of the first $$ n $$ terms of a geometric series is:

$$
S_n = a \frac{1 - r^n}{1 - r}
$$

where:
- $$ a $$ is the first term,
- $$ r $$ is the common ratio,
- $$ n $$ is the number of terms.

**Given:**
- First term $$ a = -8 $$,
- Common ratio $$ r = \frac{1}{2} $$,
- Sum $$ S_n = -\frac{63}{4} $$.

Substituting these values into the formula:

$$
-8 \cdot \frac{1 - \left( \frac{1}{2} \right)^n}{1 - \frac{1}{2}} = -\frac{63}{4}
$$

Simplify the denominator:

$$
-8 \cdot \frac{1 - \left( \frac{1}{2} \right)^n}{\frac{1}{2}} = -\frac{63}{4}
$$

Multiply by $$ 2 $$ to eliminate the fraction in the denominator:

$$
-16 \left( 1 - \left( \frac{1}{2} \right)^n \right) = -\frac{63}{4}
$$

Divide both sides by $$ -16 $$:

$$
1 - \left( \frac{1}{2} \right)^n = \frac{63}{64}
$$

Solve for $$ \left( \frac{1}{2} \right)^n $$:

$$
\left( \frac{1}{2} \right)^n = 1 - \frac{63}{64} = \frac{1}{64}
$$

Recognizing that $$ \frac{1}{64} = \left( \frac{1}{2} \right)^6 $$, we set the exponents equal:

$$
n = 6
$$

**Answer:**  
$$ n = 6 $$
$$ n = 6 $$

2.1 Determine if