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50. Uma sequência de termos é definida por \[ a_{1}=1 \quad a_{n}=(5-n) a_{n-1} \] Calcule \( \Sigma_{n=1}^{\infty} a_{n} \).

Ask by Summers Wood. in Brazil
Feb 11,2025

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A sequência começa com 1, 3, 6, 6, e depois todos os termos subsequentes são zero. Portanto, a soma infinita é 1 + 3 + 6 + 6 = 16.

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Para resolver a soma \( \Sigma_{n=1}^{\infty} a_{n} \), precisamos primeiro entender a sequência \( a_n \). Dada a definição: - \( a_{1} = 1 \) - \( a_{2} = (5-2)a_{1} = 3 \cdot 1 = 3 \) - \( a_{3} = (5-3)a_{2} = 2 \cdot 3 = 6 \) - \( a_{4} = (5-4)a_{3} = 1 \cdot 6 = 6 \) - \( a_{5} = (5-5)a_{4} = 0 \cdot 6 = 0 \) Para \( n \geq 5 \), temos que \( a_n = 0 \). Portanto, a sequência se torna: \[ a_1 = 1, \quad a_2 = 3, \quad a_3 = 6, \quad a_4 = 6, \quad a_5 = 0, \ldots \] Agora, podemos calcular a soma infinita até o quarto termo, pois os termos além de \( a_4 \) são todos zeros. \[ \Sigma_{n=1}^{\infty} a_{n} = a_1 + a_2 + a_3 + a_4 = 1 + 3 + 6 + 6 = 16 \] Portanto, a soma é: \[ \Sigma_{n=1}^{\infty} a_{n} = 16 \]

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