| Reducir: | |
| $N=\dfrac{2^{\textstyle n+4}-2^{\textstyle n+3}}{2^{\textstyle n+4}}$ | |
| $A)$ | $\dfrac{1}{2}$ |
| $B)$ | $\dfrac{1}{5}$ |
| $C)$ | $\dfrac{1}{3}$ |
| $D)$ | $2$ |
| $E)$ | $3$ |
Solución
| Resolviendo | ||
| Descomponiendo los exponentes, recordar que $n^{a+b}=n^a\cdot n^b$ | ||
| $N=\dfrac{2^{\textstyle n+4}-2^{\textstyle n+3}}{2^{\textstyle n+4}}$ | ||
| $N=\dfrac{2^n \cdot 2^4-2^n \cdot 2^3}{2^n \cdot 2^4}$ | ||
| Factorizando $2^n$ | ||
| $N=\dfrac{2^n \cdot (2^4-2^3)}{2^n (2^4)}$ | ||
| Simplificando $2^n$ | ||
| $N=\dfrac{2^4-2^3}{2^4}$ | ||
| $N=\dfrac{16-8}{16}$ | ||
| $N=\dfrac{8}{16}$ | ||
| $N=\dfrac{1}{2}$ | ||
| Respuesta: | ||
| La solución es la Alternativa A | ||