3.2 Calculate the value of \( n \) if: \( \sum_{k=1}^{n} 2(3)^{k-1}=531440 \)

To calculate the value of \( n \), we can start by simplifying the summation. The expression \( \sum_{k=1}^{n} 2(3)^{k-1} \) can be recognized as a geometric series where the first term \( a = 2 \) and the common ratio \( r = 3 \). The formula for the sum of the first \( n \) terms of a geometric series is given by: \[ S_n = a \frac{1 - r^n}{1 - r} \] Plugging in the values, we have: \[ S_n = 2 \frac{1 - 3^n}{1 - 3} = 2 \frac{1 - 3^n}{-2} = -(1 - 3^n) = 3^n - 1 \] Setting this equal to the sum provided in the problem, we have: \[ 3^n - 1 = 531440 \] Adding 1 to both sides gives: \[ 3^n = 531441 \] Now, we need to determine \( n \). Since \( 531441 \) is a power of 3, we can express it as \( 3^n \). Noting that: \[ 3^1 = 3, \quad 3^2 = 9, \quad 3^3 = 27, \quad 3^4 = 81, \quad 3^5 = 243, \quad 3^6 = 729, \quad 3^7 = 2187, \quad 3^8 = 6561, \quad 3^9 = 19683, \quad 3^{10} = 59049, \quad 3^{11} = 177147, \quad 3^{12} = 531441 \] Thus, we find \( n = 12 \). Therefore, the value of \( n \) is: \[ \boxed{12} \]