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

To determine \( n \) in the equation \( \sum_{k=1}^{n}(-8)\left(\frac{1}{2}\right)^{k-1}=-\frac{63}{4} \), we recognize that the left-hand side is a geometric series. The series can be expressed as: \[ S_n = -8 \sum_{k=1}^{n} \left(\frac{1}{2}\right)^{k-1} \] The sum of a geometric series can be calculated using the formula: \[ S_n = a \frac{1 - r^n}{1 - r} \] where \( a \) is the first term and \( r \) is the common ratio. Here, \( a = -8 \) and \( r = \frac{1}{2} \). Substituting these values, we get: \[ S_n = -8 \frac{1 - \left(\frac{1}{2}\right)^n}{1 - \frac{1}{2}} = -8 \frac{1 - \left(\frac{1}{2}\right)^n}{\frac{1}{2}} = -16 \left(1 - \left(\frac{1}{2}\right)^n\right) \] Now we set this equal to \(-\frac{63}{4}\): \[ -16 \left(1 - \left(\frac{1}{2}\right)^n\right) = -\frac{63}{4} \] Dividing both sides by -1, we have: \[ 16 \left(1 - \left(\frac{1}{2}\right)^n\right) = \frac{63}{4} \] Next, we divide both sides by 16: \[ 1 - \left(\frac{1}{2}\right)^n = \frac{63}{4 \times 16} = \frac{63}{64} \] This simplifies to: \[ 1 - \left(\frac{1}{2}\right)^n = \frac{63}{64} \] Subtracting 1 from both sides: \[ -\left(\frac{1}{2}\right)^n = \frac{63}{64} - 1 \] Calculating the right side gives: \[ \frac{63}{64} - \frac{64}{64} = -\frac{1}{64} \] Thus, we have: \[ -\left(\frac{1}{2}\right)^n = -\frac{1}{64} \] Removing the negative sign from both sides, we find: \[ \left(\frac{1}{2}\right)^n = \frac{1}{64} \] Recognizing that \(64 = 2^6\), we can rewrite the right side as: \[ \left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^6 \] Since the bases are the same, we can equate the exponents: \[ n = 6 \] Hence, the solution is: \[ \boxed{6} \]