\( \sum _ { k = 1 } ^ { m } 3 ( 2 ) ^ { k - 1 } = 93 \)

To solve the equation \( \sum _ { k = 1 } ^ { m } 3 ( 2 ) ^ { k - 1 } = 93 \), we can recognize that this is a geometric series. The sum of a geometric series can be calculated using the formula for the sum of the first \( n \) terms: \[ S_n = a \frac{1 - r^n}{1 - r} \] where \( a \) is the first term and \( r \) is the common ratio. Here, \( a = 3 \) and \( r = 2 \). Therefore, we can rewrite the sum as: \[ S_m = 3 \frac{1 - 2^m}{1 - 2} = 3(1 - 2^m)(-1) = 3(2^m - 1) \] Setting this equal to 93 gives: \[ 3(2^m - 1) = 93 \] Dividing both sides by 3, we obtain: \[ 2^m - 1 = 31 \] Adding 1 to both sides leads to: \[ 2^m = 32 \] Since \( 32 \) can be expressed as \( 2^5 \), we find that: \[ m = 5 \] Thus, the value of \( m \) is \( 5 \).