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1.5.1 What is the sum of the geometric series \( S_{n}=a+a r+a r^{2}+\ldots a r^{n-1} \) if \( r=-1 \) ?

Ask by Hill Wright. in South Africa
Mar 14,2025

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The sum of the geometric series \( S_n = a + ar + ar^2 + \ldots + ar^{n-1} \) when \( r = -1 \) is: - \( 0 \) if \( n \) is even, - \( a \) if \( n \) is odd. So, \( S_n = \frac{a(1 - (-1)^n)}{2} \).

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When the common ratio \( r = -1 \), the terms in the geometric series alternate in sign. Specifically, the series \( S_n = a + a(-1) + a(-1)^2 + \ldots + a(-1)^{n-1} \) creates a pattern where the first term is \( a \), the second term \( -a \), the third term \( a \), and so forth. If \( n \) is even, the series adds up to \( S_n = 0 \), while if \( n \) is odd, it sums to \( S_n = a \). To see this more clearly, consider an example with \( a = 1 \). The series would look like \( 1 - 1 + 1 - 1 + \ldots \). If there are 4 terms (even), it sums to \( 0 \). If there are 5 terms (odd), it sums to \( 1 \). Thus, the sum depends on whether \( n \) is even or odd!

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