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$\left(a_{n}\right)=\left(-2\dfrac {1} {2}_{,\dfrac {4} {3},\dfrac {7} {4},2}$ $\dfrac {13} {6},...$ C. $\left(a_{n}\right)=\left(0\dfrac {1} {4}_{,\dfrac {2} {3},\dfrac {9} {8},\dfrac {8} {5}}$ $\dfrac {25} {12}'...\right)$ d. {a,} $=\left(\dfrac {1} {2},\dfrac {1} {3},\dfrac {1} {5},\dfrac {1} {9},\dfrac {1} {17}1$ $\dfrac {1} {33},...\right)$ e. $\left(a_{n}\right)=\left(0\dfrac {1} {3}1^{\dfrac {2} {6}}$ 13 1' 14 8' $\dfrac {5} {27}1..\right)$ f. $\dfrac {9} {8},\dfrac {12} {11},$ 1145 ' $\left(a_{n}\right)=\left(0_{,\dfrac {3} {2},\dfrac {6} {5}\dfrac {9} {8}}$ g. {a,} $=\left(1,-\dfrac {1} {2},\dfrac {1} {3},-\dfrac {1} {4},\dfrac {1} {5}$ $-\dfrac {1} {6},..\right)$ h. {a,} $=\left(\dfrac {1} {2},-\dfrac {1} {6},$ $\dfrac {1} {12},-\dfrac {1} {20},\dfrac {1} {30},..\right)$ i. $\left(a_{n}\right)=\left(\dfrac {1} {5^{3}}_{,\dfrac {3} {5^{5}}}$ $\dfrac {5} {5^{7}},\dfrac {7} {5^{9}},$ 59'. 1 ' j. $\left(a_{n}\right)=$ $=\left(\dfrac {1} {2},-\dfrac {1} {4},$ 61 ! $-\dfrac {1} {8},$ 11 0 k.
Bachillerato
Aritmética y álgebra
Búsquedas: 125
Contenido de la pregunta
encontrar el término general de cada una de las siguientes sucesiones
Solución
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Profesor de Qanda - Gabriela
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Estimada Alejandra, no dude en realizar cualquier pregunta aclaratoria. Muchas gracias!