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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 \( r=-1 \) ?

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Mar 10,2025

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The sum of the geometric series \( S_n = a + ar + ar^2 + \ldots + ar^{n-1} \) when \( r = -1 \) is: \[ S_n = \frac{a(1 - (-1)^n)}{2} \]

Solución

To find the sum of a geometric series, we can use the formula for the sum of a finite geometric series: \[ S_n = \frac{a(1 - r^n)}{1 - r} \] where: - \( S_n \) is the sum of the first \( n \) terms of the series - \( a \) is the first term of the series - \( r \) is the common ratio of the series - \( n \) is the number of terms in the series Given that \( r = -1 \), we can substitute this value into the formula to find the sum of the series. Let's calculate the sum of the geometric series with \( r = -1 \). Simplify the expression by following steps: - step0: Solution: \(\frac{a\left(1-\left(-1\right)^{n}\right)}{\left(1-\left(-1\right)\right)}\) - step1: Remove the parentheses: \(\frac{a\left(1-\left(-1\right)^{n}\right)}{1-\left(-1\right)}\) - step2: Remove the parentheses: \(\frac{a\left(1-\left(-1\right)^{n}\right)}{1+1}\) - step3: Multiply the terms: \(\frac{a-a\left(-1\right)^{n}}{1+1}\) - step4: Add the numbers: \(\frac{a-a\left(-1\right)^{n}}{2}\) The sum of the geometric series \( S_n = a + ar + ar^2 + \ldots + ar^{n-1} \) when \( r = -1 \) is given by the formula: \[ S_n = \frac{a(1 - (-1)^n)}{2} \] This formula provides the sum of the first \( n \) terms of the geometric series with a common ratio of \( -1 \).

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Bonus Knowledge

When the common ratio \( r = -1 \), the terms of the geometric series alternate between \( a \) and \( -a \). The series can be expressed as \( S_n = a - a + a - a + \ldots \). For even \( n \), the sum \( S_n \) will equal \( 0 \) as the positive and negative terms cancel each other out. However, for odd \( n \), there will be one extra \( a \), making the sum \( S_n = a \). For example, if \( n = 5 \) (which is odd), the series would look like \( a - a + a - a + a = a \). Conversely, for \( n = 4 \) (even), it turns into \( a - a + a - a = 0 \). Thus, the final answer depends on \( n \) being odd or even!

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