\( \sum _ { i = 2 } ^ { m } 32 ( 2 ) ^ { 4 - i } < 250 \)

To solve the inequality \( \sum _ { i = 2 } ^ { m } 32 ( 2 ) ^ { 4 - i } < 250 \), let's first simplify the sum: Notice that \( 32(2)^{4 - i} \) can be rewritten for each \( i \): - For \( i=2 \): \( 32(2)^{4-2} = 32(2^2) = 32 \cdot 4 = 128 \) - For \( i=3 \): \( 32(2)^{4-3} = 32(2^1) = 32 \cdot 2 = 64 \) - For \( i=4 \): \( 32(2)^{4-4} = 32(2^0) = 32 \cdot 1 = 32 \) Continuing this, we see a pattern: \[ \sum_{i=2}^m 32(2)^{4-i} = 32 \left( 2^2 + 2^1 + 2^0 + \ldots + 2^{4-m} \right) \] This is a finite geometric series where the ratio is \( \frac{1}{2} \) up to \( m \) terms, starting from \( \frac{32}{4} \) when \( i=2 \) down to \( \frac{32}{2^{m-1}} \) when \( m \) is increased. Now we can rewrite the sum: \[ S = 32 \sum_{k=0}^{m-2} (2^k) \] Using the formula for the sum of a geometric series \( S = a \frac{1-r^n}{1-r} \) where \( a \) is the first term and \( r \) is the common ratio: \[ = 32 \cdot \frac{2^0 (1 - (1/2)^{m-1})}{1 - 1/2} \] \[ = 32 \cdot \frac{(1 - (1/2)^{m-1})}{(1/2)} \] \[ = 64 (1 - (1/2)^{m-1}) \] Setting up our inequality: \[ 64 (1 - (1/2)^{m-1}) < 250 \] \[ 1 - (1/2)^{m-1} < \frac{250}{64} = 3.90625 \] Subtract \( 1 \) and multiply by \(-1\) we get: \[ (1/2)^{m-1} > 1 - \frac{250}{64} = -2.90625 \] Since the left side \( (1/2)^{m-1} \) is always positive, and \( -2.90625 \) is a negative number, therefore this condition is satisfied. To find the maximum \( m \), set up: \[ 64 (1 - (1/2)^{m-1}) < 250 \rightarrow (1/2)^{m-1} < \frac{64}{250} \] \[ (1/2)^{m-1} < \frac{64}{250} \] Determine its logarithm or successively checking values of \( m \). Hence m= 6 is the valid integer satisfying the original condition: \[ \sum _ { i = 2 } ^ { 6 } 32 ( 2 ) ^ { 4 - i } < 250 \] Thus, the solution to \( \sum _ { i = 2 } ^ { m } 32 ( 2 ) ^ { 4 - i } < 250 \) is \( m = 6 \).