b) \( \sum_{i=1}^{n} 3 \cdot 2^{i-1}=12285 \)
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To solve the equation \( \sum_{i=1}^{n} 3 \cdot 2^{i-1}=12285 \), we can first simplify the summation. It's a geometric series, where the first term \( a = 3 \) and the common ratio \( r = 2 \). The sum of the first \( n \) terms of a geometric series can be expressed as: \[ S_n = a \frac{r^n - 1}{r - 1} \] Plugging in our values: \[ S_n = 3 \frac{2^n - 1}{2 - 1} = 3(2^n - 1) \] Setting it equal to 12285: \[ 3(2^n - 1) = 12285 \] Dividing both sides by 3 gives: \[ 2^n - 1 = 4095 \] Adding 1 to both sides: \[ 2^n = 4096 \] Now, recognizing that \( 4096 = 2^{12} \), we find: \[ n = 12 \] Thus, the value of \( n \) is \( \boxed{12} \).