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Find the sum \( \sum_{n=1}^{12} 5 \cdot 5^{n-1}= \)

Ask by Montgomery Crawford. in the United States
Mar 08,2025

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The sum \( \sum_{n=1}^{12} 5 \cdot 5^{n-1} \) is 305,175,780.

Solución

To find the sum \( \sum_{n=1}^{12} 5 \cdot 5^{n-1} \), we can first recognize that this is a geometric series. The general form of a geometric series is given by: \[ S = a \cdot \frac{1 - r^n}{1 - r} \] where: - \( S \) is the sum of the series, - \( a \) is the first term, - \( r \) is the common ratio, - \( n \) is the number of terms. In our case: - The first term \( a = 5 \cdot 5^{1-1} = 5 \cdot 1 = 5 \), - The common ratio \( r = 5 \), - The number of terms \( n = 12 \). Now we can substitute these values into the formula: \[ S = 5 \cdot \frac{1 - 5^{12}}{1 - 5} \] Calculating the denominator: \[ 1 - 5 = -4 \] Now substituting back into the sum formula: \[ S = 5 \cdot \frac{1 - 5^{12}}{-4} \] This simplifies to: \[ S = -\frac{5}{4} (1 - 5^{12}) \] Now we need to calculate \( 5^{12} \). Let's compute that first. Calculating \( 5^{12} \): \[ 5^{12} = 244140625 \] Now substituting this value back into the sum: \[ S = -\frac{5}{4} (1 - 244140625) = -\frac{5}{4} (-244140624) = \frac{5 \cdot 244140624}{4} \] Now we can calculate \( \frac{5 \cdot 244140624}{4} \): \[ 5 \cdot 244140624 = 1220703120 \] Now dividing by 4: \[ \frac{1220703120}{4} = 305175780 \] Thus, the sum \( \sum_{n=1}^{12} 5 \cdot 5^{n-1} \) is: \[ \boxed{305175780} \]

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Let's simplify the given sum \( \sum_{n=1}^{12} 5 \cdot 5^{n-1} \). We can factor out the constant \( 5 \): \[ \sum_{n=1}^{12} 5 \cdot 5^{n-1} = 5 \sum_{n=1}^{12} 5^{n-1} \] Next, we notice that the sum \( \sum_{n=1}^{12} 5^{n-1} \) is a geometric series. The first term of the series is \( a = 5^0 = 1 \) (when \( n=1 \)), and the common ratio is \( r = 5 \). The formula for the sum of the first \( m \) terms of a geometric series is given by: \[ S_m = a \frac{r^m - 1}{r - 1} \] In our case, \( m = 12 \), \( a = 1 \), and \( r = 5 \): \[ S_{12} = 1 \cdot \frac{5^{12} - 1}{5 - 1} = \frac{5^{12} - 1}{4} \] Now substituting this back into our expression for the original sum: \[ \sum_{n=1}^{12} 5 \cdot 5^{n-1} = 5 \cdot \frac{5^{12} - 1}{4} \] Simplifying further: \[ = \frac{5(5^{12} - 1)}{4} \] Calculating \( 5(5^{12} - 1) \): \[ = 5^{13} - 5 \] Thus, we have: \[ \sum_{n=1}^{12} 5 \cdot 5^{n-1} = \frac{5^{13} - 5}{4} \] Therefore, the final answer for the sum is: \[ \boxed{\frac{5^{13} - 5}{4}} \]

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