Find the sum \( \sum_{n=1}^{12} 5 \cdot 5^{n-1}= \)
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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: \[ \boxed{\frac{5^{13} - 5}{4}} \]
