To find the sum of the first 12 terms of the geometric series \(3 + 6 + 12 + 24 + \ldots\), we first need to identify the first term \(a\) and the common ratio \(r\).
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Identify the first term and common ratio:
- The first term \(a = 3\).
- The common ratio \(r\) can be found by dividing the second term by the first term: \[ r = \frac{6}{3} = 2. \]
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Formula for the sum of the first \(n\) terms of a geometric series: The sum \(S_n\) of the first \(n\) terms of a geometric series can be calculated using the formula: \[ S_n = a \frac{1 - r^n}{1 - r} \] where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms.
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Plug in the values for \(S_{12}\): For this series, we have \(a = 3\), \(r = 2\), and \(n = 12\). \[ S_{12} = 3 \frac{1 - 2^{12}}{1 - 2} \] The denominator simplifies to: \[ 1 - 2 = -1. \] Therefore, the formula simplifies to: \[ S_{12} = 3 \frac{1 - 2^{12}}{-1} = -3 (1 - 2^{12}) = 3 (2^{12} - 1). \]
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Calculate \(2^{12}\): \[ 2^{12} = 4096. \]
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Substitute this value back into the formula: \[ S_{12} = 3 (4096 - 1) = 3 (4095) = 12285. \]
Thus, the sum of the first 12 terms is: \[ \boxed{12285}. \]
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