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4. Calcular S dados: \( \begin{array}{lll}\text { a. } a_{1}=2 ; r=3 ; n=6 . & \text { c) } a_{1}=8 ; r=5 ; n=4 & \text { e) } a_{1}=1215 ; r=1 / 3 ; n=6 \\ \text { b. } a_{1}=4 ; r=6 ; a_{n}=31104 . & \text { d) } A_{1}=243 ; r=r=3 / 8 ; n=6 . & \end{array} \)

Ask by Ortiz Bush. in Bolivia
Mar 11,2025

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Los resultados de las sumas de las series son: - **a)** \( S = 728 \) - **b)** \( S = 7776 \) - **c)** \( S = 1248 \) - **d)** \( S \approx 387.72 \) - **e)** \( S = 1820 \)

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Para calcular la suma de los términos de una progresión geométrica, puedes usar la fórmula: \[ S_n = a_1 \cdot \frac{1 - r^n}{1 - r} \quad \text{(si } r \neq 1\text{)} \] Para cada uno de los conjuntos de datos, primero vamos a calcular el valor de \(S_n\): a) \( a_{1} = 2 \), \( r = 3 \), \( n = 6 \) \[ S_6 = 2 \cdot \frac{1 - 3^6}{1 - 3} = 2 \cdot \frac{1 - 729}{-2} = 2 \cdot \frac{-728}{-2} = 728 \] b) Aquí debemos encontrar \(a_{n}\): \[ a_n = a_1 \cdot r^{n-1} \Rightarrow 31104 = 4 \cdot r^{n-1} \] \[ r^{n-1} = \frac{31104}{4} = 7776 \Rightarrow \sqrt[5]{7776} = r = 6 \] c) \( a_{1} = 8 \), \( r = 5 \), \( n = 4 \) \[ S_4 = 8 \cdot \frac{1 - 5^4}{1 - 5} = 8 \cdot \frac{1 - 625}{-4} = 8 \cdot \frac{-624}{-4} = 1248 \] d) \( a_{1} = 243, \ r = \frac{3}{8}, \ n = 6 \) \[ S_6 = 243 \cdot \frac{1 - \left(\frac{3}{8}\right)^6}{1 - \frac{3}{8}} = 243 \cdot \frac{1 - \frac{729}{262144}}{\frac{5}{8}} \approx 243 \cdot 4.424 \approx 1076.87 \] e) \( a_{1} = 1215, \ r = \frac{1}{3}, \ n = 6 \) \[ S_6 = 1215 \cdot \frac{1 - \left(\frac{1}{3}\right)^6}{1 - \frac{1}{3}} = 1215 \cdot \frac{1 - \frac{1}{729}}{\frac{2}{3}} = 1215 \cdot \frac{728/729}{2/3} \approx 1215 \cdot 1.095 \approx 1333.74 \] Los resultados finales son: a) \( S_6 = 728 \) b) \( r = 6 \) c) \( S_4 = 1248 \) d) \( S_6 \approx 1076.87 \) e) \( S_6 \approx 1333.74 \)

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