\( \sum _ { p = 0 } ^ { m } 3 ( 2 ) ^ { 2 p - 3 } < 8000 \)

To solve the inequality: \[ \sum_{p=0}^{m} 3 \cdot 2^{2p - 3} < 8000 \] **Step 1: Simplify the Sum** First, simplify the expression inside the summation: \[ 3 \cdot 2^{2p - 3} = 3 \cdot \frac{2^{2p}}{2^3} = \frac{3}{8} \cdot 4^p \] So, the sum becomes: \[ \sum_{p=0}^{m} \frac{3}{8} \cdot 4^p = \frac{3}{8} \sum_{p=0}^{m} 4^p \] **Step 2: Use the Geometric Series Formula** The sum of a geometric series \(\sum_{p=0}^{m} ar^p\) is given by: \[ \sum_{p=0}^{m} ar^p = a \cdot \frac{r^{m+1} - 1}{r - 1} \] In this case, \(a = \frac{3}{8}\) and \(r = 4\): \[ \frac{3}{8} \cdot \frac{4^{m+1} - 1}{4 - 1} = \frac{4^{m+1} - 1}{8} \] **Step 3: Set Up the Inequality** Set the simplified sum less than 8000: \[ \frac{4^{m+1} - 1}{8} < 8000 \] Multiply both sides by 8: \[ 4^{m+1} - 1 < 64000 \] Add 1 to both sides: \[ 4^{m+1} < 64001 \] **Step 4: Solve for \(m\)** Find the largest integer \(m\) such that \(4^{m+1} < 64001\). Calculate powers of 4: \[ 4^7 = 16384 \quad \text{and} \quad 4^8 = 65536 \] Since \(4^8 = 65536 > 64001\), the largest integer \(m+1\) can be is 7. Therefore: \[ m + 1 \leq 7 \implies m \leq 6 \] **Verification:** - For \(m = 6\): \[ \frac{4^{7} - 1}{8} = \frac{16384 - 1}{8} = \frac{16383}{8} = 2047.875 < 8000 \] - For \(m = 7\): \[ \frac{4^{8} - 1}{8} = \frac{65536 - 1}{8} = \frac{65535}{8} = 8191.875 > 8000 \] Thus, the largest integer \(m\) satisfying the inequality is **6**. **Answer:** \( m = 6 \)