Determine the value of \( p \) in the following series: \( \sum_{i=2}^{p} 3^{i-1}=1092 \)

To determine the value of \( p \) in the series \( \sum_{i=2}^{p} 3^{i-1} = 1092 \), we can start by simplifying the sum. The series can be re-indexed. Notice that when \( i \) starts at 2, the term \( 3^{i-1} \) starts at \( 3^{1} \). Therefore, we rewrite the sum: \[ \sum_{i=2}^{p} 3^{i-1} = 3^{1} + 3^{2} + 3^{3} + \ldots + 3^{p-1} \] This is a geometric series with the first term \( a = 3 \) and the common ratio \( r = 3 \). The number of terms in the series from \( i=2 \) to \( p \) is \( p - 1 \). The formula for the sum \( S_n \) of the first \( n \) terms of a geometric series is given by: \[ S_n = a \frac{r^n - 1}{r - 1} \] Applying this formula, we have: - \( a = 3 \) - \( r = 3 \) - \( n = p - 1 \) Thus, the sum becomes: \[ \sum_{i=2}^{p} 3^{i-1} = 3 \frac{3^{p-1} - 1}{3 - 1} = 3 \frac{3^{p-1} - 1}{2} \] Setting this equal to 1092: \[ 3 \frac{3^{p-1} - 1}{2} = 1092 \] Multiplying through by 2 to eliminate the fraction yields: \[ 3(3^{p-1} - 1) = 2184 \] Dividing by 3 gives: \[ 3^{p-1} - 1 = 728 \] Adding 1 to both sides, we have: \[ 3^{p-1} = 729 \] Recognizing that \( 729 \) is a power of \( 3 \): \[ 729 = 3^6 \] Thus, we can set the exponents equal to each other: \[ p - 1 = 6 \] Solving for \( p \): \[ p = 7 \] Therefore, the value of \( p \) is \( \boxed{7} \).