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

To solve the equation: \[ \sum _ { k = 1 } ^ { m } 3 \cdot 2^{k - 1} = 93 \] we'll follow these steps: 1. **Recognize the Type of Series:** The given sum is a *geometric series*, where each term is obtained by multiplying the previous term by a common ratio. 2. **Factor Out Constants:** \[ \sum _ { k = 1 } ^ { m } 3 \cdot 2^{k - 1} = 3 \cdot \sum _ { k = 1 } ^ { m } 2^{k - 1} \] 3. **Use the Formula for the Sum of a Geometric Series:** For a geometric series with first term \( a \) and common ratio \( r \), the sum of \( m \) terms is: \[ S = a \cdot \frac{r^m - 1}{r - 1} \] In this case: - First term \( a = 2^{0} = 1 \) - Common ratio \( r = 2 \) So, \[ \sum _ { k = 1 } ^ { m } 2^{k - 1} = \frac{2^{m} - 1}{2 - 1} = 2^{m} - 1 \] 4. **Substitute Back into the Original Equation:** \[ 3 \cdot (2^{m} - 1) = 93 \] 5. **Solve for \( m \):** - Divide both sides by 3: \[ 2^{m} - 1 = \frac{93}{3} = 31 \] - Add 1 to both sides: \[ 2^{m} = 32 \] - Recognize that \( 32 \) is a power of 2: \[ 32 = 2^5 \] - Therefore: \[ 2^{m} = 2^{5} \implies m = 5 \] **Conclusion:** The value of \( m \) that satisfies the equation is **5**.