Exercício sobre álgebra elementar

Sendo $\phantom{X}n\phantom{X}$ um número natural ($\;n\,\in\,\mathbb{N}\;$), mostre que $\;\dfrac{2^{\large n}\,+\,2^{\large n+1}\,+\,2^{\large n+2}}{2^{\large n+3}\,+\,2^{\large n+4}}\,=\,\dfrac{7}{24}\;$.

resposta:

Resolução:$\;\dfrac{2^{\large n}\,+\,2^{\large n+1}\,+\,2^{\large n+2}}{2^{\large n+3}\,+\,2^{\large n+4}}\,=\,\dfrac{2^{\large n}\,+\,2^{\large n}\centerdot 2\,+\,2^{\large n}\centerdot 2^{\large 2}}{2^{\large n}\centerdot 2^{\large 3}\,+\,2^{\large n}\centerdot 2^{\large 4}}\,$ $=\;\dfrac{2^{\large n}\centerdot(1\,+\,2^{\large 1}\,+\,2^{\large 2})}{2^{\large n}\centerdot(2^{\large 3}\,+\,2^{\large 4})}\,=\,\dfrac{2^{\large n}\centerdot 7}{2^{\large n}\centerdot 24}\,=\,\dfrac{7}{24}\;$.