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Calculate \( p \) if \( \sum_{n=1}^{\infty} 27 p^{2}=\sum_{k=2}^{5}\left(1-t^{2}\right) \)

Ask by Weber Chambers. in South Africa
Mar 09,2025

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Cannot determine the value of \( p \) without knowing the value of \( t \).

Solución

To solve the given equation, we need to calculate the sum of the series on both sides of the equation. Given equation: \[ \sum_{n=1}^{\infty} 27 p^{2} = \sum_{k=2}^{5}\left(1-t^{2}\right) \] Let's calculate the sum of the series on both sides: For the left side: \[ \sum_{n=1}^{\infty} 27 p^{2} = 27 p^{2} + 27 p^{2} + 27 p^{2} + \ldots \] This is an infinite geometric series with first term \(27 p^{2}\) and common ratio \(p^{2}\). The sum of an infinite geometric series is given by the formula: \[ S = \frac{a}{1 - r} \] where \(a\) is the first term and \(r\) is the common ratio. Substitute the values into the formula: \[ S = \frac{27 p^{2}}{1 - p^{2}} \] For the right side: \[ \sum_{k=2}^{5}\left(1-t^{2}\right) = (1-t^{2}) + (1-t^{2}) + (1-t^{2}) + (1-t^{2}) \] This is a finite series with four terms. The sum of a finite series is the sum of the individual terms. Substitute the values into the formula: \[ S = 4(1-t^{2}) \] Now, we can equate the two sums: \[ \frac{27 p^{2}}{1 - p^{2}} = 4(1-t^{2}) \] Solve for \(p\). Solve the equation by following steps: - step0: Solve for \(p\): \(\frac{27p^{2}}{1-p^{2}}=4\left(1-t^{2}\right)\) - step1: Rewrite the expression: \(\frac{27p^{2}}{1-p^{2}}=4-4t^{2}\) - step2: Multiply both sides of the equation by LCD: \(\frac{27p^{2}}{1-p^{2}}\times \left(1-p^{2}\right)=\left(4-4t^{2}\right)\left(1-p^{2}\right)\) - step3: Simplify the equation: \(27p^{2}=4-4t^{2}+\left(-4+4t^{2}\right)p^{2}\) - step4: Move the expression to the left side: \(27p^{2}-\left(4-4t^{2}+\left(-4+4t^{2}\right)p^{2}\right)=0\) - step5: Remove the parentheses: \(27p^{2}-4+4t^{2}+\left(4-4t^{2}\right)p^{2}=0\) - step6: Simplify: \(\left(31-4t^{2}\right)p^{2}-4+4t^{2}=0\) - step7: Move the constant to the right side: \(\left(31-4t^{2}\right)p^{2}=4-4t^{2}\) - step8: Divide both sides: \(\frac{\left(31-4t^{2}\right)p^{2}}{31-4t^{2}}=\frac{4-4t^{2}}{31-4t^{2}}\) - step9: Divide the numbers: \(p^{2}=\frac{4-4t^{2}}{31-4t^{2}}\) - step10: Simplify the expression: \(p=\pm \sqrt{\frac{4-4t^{2}}{31-4t^{2}}}\) - step11: Simplify: \(p=\pm \frac{2\sqrt{31-35t^{2}+4t^{4}}}{31-4t^{2}}\) - step12: Separate into possible cases: \(\begin{align}&p=\frac{2\sqrt{31-35t^{2}+4t^{4}}}{31-4t^{2}}\\&p=-\frac{2\sqrt{31-35t^{2}+4t^{4}}}{31-4t^{2}}\end{align}\) The solutions for \(p\) are: \[ p = \frac{2\sqrt{31-35t^{2}+4t^{4}}}{31-4t^{2}} \] \[ p = -\frac{2\sqrt{31-35t^{2}+4t^{4}}}{31-4t^{2}} \] Please note that these solutions are in terms of \(t\), which is not defined in the given equation. Therefore, we cannot determine a specific value for \(p\) without knowing the value of \(t\).

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To solve for \( p \) in the equation \( \sum_{n=1}^{\infty} 27 p^{2} = \sum_{k=2}^{5}(1-t^{2}) \), we first analyze each side of the equation. 1. **Left Side:** The left side is a series defined as \( \sum_{n=1}^{\infty} 27 p^{2} \). This is a constant series where the term \( 27 p^{2} \) is repeated infinitely. Therefore, it can be interpreted as \( 27 p^{2} \) multiplied by the number of terms. Since the series diverges for \( p \neq 0 \), we need to clarify the context regarding convergence. If the series were to converge, then \( p \) must be strictly less than 1 (i.e., \( |p| < 1 \)), but there are nuances here. For our case, we assume: \[ \sum_{n=1}^{\infty} 27 p^{2} = 27 p^{2} \cdot \infty \quad (\text{diverges if } p \neq 0) \] This means that the left-hand side diverges as long as \( p \) is not 0. 2. **Right Side:** The right side \( \sum_{k=2}^{5}(1-t^{2}) \) consists of evaluating the sum of a specific term. Notice we can sum it directly by evaluating \( (1 - t^{2}) \) at \( k=2, 3, 4, 5 \): The right-hand side computes as: \[ (1-t^{2}) + (1-t^{2}) + (1-t^{2}) + (1-t^{2}) = 4(1-t^{2}) \] Thus, the right-hand side simplifies to \( 4(1 - t^{2}) \). Equating both sides: \[ 27 p^{2} (\text{diverging}) = 4(1 - t^{2}) \] For equality to hold, considering the divergence of the left side, it implies \( 0 = 4(1 - t^2) \). Therefore: \[ 1 - t^{2} = 0 \implies t^{2} = 1 \implies t = \pm 1 \] However, we want to focus on finding a value for \( p \). The divergence suggests that we explore \( p = 0 \) too. If \( p = 0\): - The left side becomes \( 0 \rightarrow 27(0) = 0 \). - The right side confirms \( 4(1 - t^{2}) = 4(0) = 0 \) (assuming \( t^{2} = 1 \)). Given this analysis, a suitable solution is that: \[ \text{If } p = 0, \text{ both sides yield } 0. \] As a conclusion: \[ \boxed{0} \] Thus, the correct value of \( p \) is \( 0 \).

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