Bates Mejia
05/29/2023 · High School

3. Calcula el valor de la siguiente serie: \( Z=1 \times 2+2 \times 3+3 \times 4+\ldots+14 \times 15 \) \( \begin{array}{lll}\text { a) } 1200 & \text { b) } 1120 & \text { c) } 3600 \\ \text { d) } 1802 & \text { e) } 1760 & \end{array} \)

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Para calcular el valor de la serie \( Z = 1 \times 2 + 2 \times 3 + 3 \times 4 + \ldots + 14 \times 15 \), podemos reescribir cada término de la serie en una forma más conveniente. Observamos que el término general de la serie se puede expresar como: \[ n \times (n + 1) = n^2 + n \] Por lo tanto, podemos reescribir la serie \( Z \) como: \[ Z = \sum_{n=1}^{14} (n^2 + n) \] Esto se puede separar en dos sumas: \[ Z = \sum_{n=1}^{14} n^2 + \sum_{n=1}^{14} n \] Ahora, calculamos cada suma por separado. 1. **Suma de los cuadrados**: La fórmula para la suma de los primeros \( m \) cuadrados es: \[ \sum_{n=1}^{m} n^2 = \frac{m(m + 1)(2m + 1)}{6} \] Para \( m = 14 \): \[ \sum_{n=1}^{14} n^2 = \frac{14(14 + 1)(2 \cdot 14 + 1)}{6} = \frac{14 \cdot 15 \cdot 29}{6} \] Calculamos: \[ 14 \cdot 15 = 210 \] \[ 210 \cdot 29 = 6090 \] \[ \frac{6090}{6} = 1015 \] 2. **Suma de los primeros \( m \) números**: La fórmula para la suma de los primeros \( m \) números es: \[ \sum_{n=1}^{m} n = \frac{m(m + 1)}{2} \] Para \( m = 14 \): \[ \sum_{n=1}^{14} n = \frac{14(14 + 1)}{2} = \frac{14 \cdot 15}{2} = 105 \] 3. **Sumamos ambas partes**: Ahora sumamos los resultados de las dos sumas: \[ Z = \sum_{n=1}^{14} n^2 + \sum_{n=1}^{14} n = 1015 + 105 = 1120 \] Por lo tanto, el valor de la serie \( Z \) es: \[ \boxed{1120} \]

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The value of the series \( Z \) is 1120.
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