Question

aratati ca:2 la puterea n+3 +2 la puterea n+2 +2 la puterea n +1 + 2la puterea n/3la puterea n +2 +2*3 la puterea n+1+3la puterea n se simplifica prin6​

Answer 1

$\dfrac{2^{n+3} + 2^{n+2} + 2^{n+1} + 2^{n}}{3^{n+2} + 2\cdot 3^{n+1} + 3^n}\\ \\= \dfrac{2^n\Big(2^3+2^2+2+1\Big)}{3^n\Big(3^2 + 2\cdot 3 + 1\Big)} \\ \\ = \dfrac{2^n\cdot 15}{3^n \Big(9 + 6 + 1\Big)} = \dfrac{2^{n-1}\cdot 30}{3^n\cdot 16} = \dfrac{2^{n-1}\cdot 30}{3^{n-1} \cdot 48}\\ \\ 6 \mid 30 \text{ si } 6 \mid 48 \implies \text{Fractia se poate simplifica prin 6},(\forall)n\in\mathbb{N}^*$

Answer 2

$\dfrac{2^{n+3}+2^{n+2}+2^{n+1}+2^{n}}{3^{n+2}+2\cdot 3^{n+1}+3^{n}} = \dfrac{2^n\cdot(2^3+2^2+2^1+1)}{3^n\cdot (3^2+2\cdot 3+1)} = \\ \\\\ = \dfrac{2^n\cdot 15}{3^n\cdot 16} = \dfrac{2^{n-1}\cdot 2\cdot 15}{3^{n-1}\cdot 3\cdot 16} = \dfrac{2^{n-1}\cdot 6\cdot 5}{3^{n-1}\cdot 6\cdot 8}\\ \\ \\\text{Se simplifica cu 6, }\forall n\in \mathbb{N}^*$