\( 2.2 \quad \) Given: \( \sum_{p=k}^{10} 3^{p-1}=29520 \). Calculate the value of \( k \).

To solve the equation \( \sum_{p=k}^{10} 3^{p-1} = 29520 \), we first rewrite the summation: \[ \sum_{p=k}^{10} 3^{p-1} = 3^{k-1} + 3^k + 3^{k+1} + \ldots + 3^{9} + 3^{10} \] This can be recognized as a geometric series. The first term of the series is \( 3^{k-1} \) and the last term is \( 3^{9} \). The number of terms in this series is \( 10 - k + 1 = 11 - k \). The sum of a geometric series can be calculated using the formula: \[ S_n = a \frac{r^n - 1}{r - 1} \] where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms. Here, \( a = 3^{k-1} \), \( r = 3 \), and \( n = 11 - k \). Thus, the sum becomes: \[ \sum_{p=k}^{10} 3^{p-1} = 3^{k-1} \frac{3^{11 - k} - 1}{3 - 1} = 3^{k-1} \frac{3^{11 - k} - 1}{2} \] Setting this equal to \( 29520 \): \[ 3^{k-1} \frac{3^{11 - k} - 1}{2} = 29520 \] Multiplying both sides by 2 gives: \[ 3^{k-1} (3^{11 - k} - 1) = 59040 \] Now, we can simplify \( 3^{k-1} (3^{11 - k} - 1) \): \[ 3^{k-1} (3^{11 - k} - 1) = 3^{10} - 3^{k-1} = 59040 \] Calculating \( 3^{10} \): \[ 3^{10} = 59049 \] Therefore, we can now rewrite the equation: \[ 59049 - 3^{k-1} = 59040 \] This leads us to: \[ 3^{k-1} = 59049 - 59040 = 9 \] Since \( 9 = 3^2 \): \[ k - 1 = 2 \quad \Rightarrow \quad k = 3 \] Thus, the value of \( k \) is \( \boxed{3} \).