Mcfarlane Tran
11/22/2023 · Elementary School

2. Calcula utilizando propiedades: \[ \sqrt{\left(6.5^{n-1}+20.5^{n-2}-25.5^{n-3}\right)+\left(\frac{5^{n+1}}{5^{2}}\right)} \]

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Para simplificar la expresión dada, primero reescribimos la parte dentro de la raíz cuadrada: \[ \sqrt{6.5^{n-1} + 20.5^{n-2} - 25.5^{n-3} + \frac{5^{n+1}}{5^2}} \] Observamos que \(\frac{5^{n+1}}{5^2} = 5^{n+1-2} = 5^{n-1}\). Entonces, podemos reescribir la expresión como: \[ \sqrt{6.5^{n-1} + 20.5^{n-2} - 25.5^{n-3} + 5^{n-1}} \] Ahora, agrupamos los términos que tienen la misma base: \[ \sqrt{(6 + 5)5^{n-1} + 20.5^{n-2} - 25.5^{n-3}} \] Esto se simplifica a: \[ \sqrt{11 \cdot 5^{n-1} + 20.5^{n-2} - 25.5^{n-3}} \] Ahora, para simplificar aún más, podemos factorizar \(5^{n-3}\) de los términos que lo contienen: \[ \sqrt{11 \cdot 5^{n-1} + 20 \cdot 5^{n-2} - 25} \] Factorizamos \(5^{n-3}\): \[ = \sqrt{5^{n-3} \left(11 \cdot 5^2 + 20 \cdot 5 - 25\right)} \] Calculamos \(11 \cdot 5^2 + 20 \cdot 5 - 25\): \[ = 11 \cdot 25 + 100 - 25 = 275 + 100 - 25 = 350 \] Por lo tanto, la expresión se convierte en: \[ \sqrt{5^{n-3} \cdot 350} \] Utilizando la propiedad de la raíz cuadrada, podemos separar: \[ = \sqrt{5^{n-3}} \cdot \sqrt{350} \] Esto se simplifica a: \[ = 5^{\frac{n-3}{2}} \cdot \sqrt{350} \] Finalmente, podemos expresar \(\sqrt{350}\) como: \[ \sqrt{350} = \sqrt{25 \cdot 14} = 5\sqrt{14} \] Por lo tanto, la expresión final es: \[ = 5^{\frac{n-3}{2}} \cdot 5 \sqrt{14} = 5^{\frac{n-1}{2}} \sqrt{14} \] Así que la respuesta final es: \[ \boxed{5^{\frac{n-1}{2}} \sqrt{14}} \]

Quick Answer

La respuesta final es \(5^{\frac{n-1}{2}} \sqrt{14}\).
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