8.2 Determine the value of P if \( \mathrm{P}=\sum_{k=1}^{13} 3^{k-5} \)

To determine the value of \( P = \sum_{k=1}^{13} 3^{k-5} \), we can first simplify the summation: \[ P = \sum_{k=1}^{13} 3^{k-5} = \sum_{k=1}^{13} \frac{3^k}{3^5} = \frac{1}{3^5} \sum_{k=1}^{13} 3^k \] Now, we recognize that \( \sum_{k=1}^{n} a^k \) for a geometric series can be computed using the formula: \[ \sum_{k=1}^{n} a^k = a \frac{a^n - 1}{a - 1} \] In our case, \( a = 3 \) and \( n = 13 \): \[ \sum_{k=1}^{13} 3^k = 3 \frac{3^{13} - 1}{3 - 1} = 3 \frac{3^{13} - 1}{2} = \frac{3^{14} - 3}{2} \] Now substituting this back into our equation for \( P \): \[ P = \frac{1}{3^5} \cdot \frac{3^{14} - 3}{2} = \frac{3^{14} - 3}{2 \cdot 3^5} \] Now simplifying this further, \( 3^{14 - 5} = 3^9 \): \[ P = \frac{3^9 - 1.5}{2} = \frac{3^9}{2} - \frac{1.5}{2} = \frac{3^9}{2} - 0.75 \] In numeric terms, compute \( 3^9 \): \[ 3^9 = 19683 \] So, \[ P = \frac{19683}{2} - 0.75 = 9841.5 - 0.75 = 9840.75 \] However, integrating back to our sum structure and maintaining proper volume and whole associated variables states: \[ P = 3^{-5} \cdot \frac{3^{14} - 3}{2} = \frac{3^9 - 1.5}{2} \rightarrow \frac{19683 - 3}{2 \cdot 243} \] Summarizing towards finality, thus in pure progression, all yields converge fully: \[ P = \frac{19680}{2 \cdot 243} = \frac{9840}{243} \] Thus, the clean resolution yields: \[ P = 9840 \] Summarizing: The computed value of \( P \) in conclusion stands as \( 9840 \).