Question

4^n. 2^(n-1) / 4^(n+1)

Answer 1

vamos lá

$\frac{4^n.2^n^-^1}{4^n^+^1} = \\ \\ \frac{ 2^{2n}. 2^{n-1} }{ 2^{2(n+1)} } = \\ \\ \frac{ 2^{2n+n-1} }{ 2^{2n+2} } = \\ \\ \frac{ 2^{3n-1} }{ 2^{2n+2} } = \\ \\ 2^{3n-1-(2n+2)} = \\ \\ 2^{3n-1-2n-2} = \\ \\ 2^{n-3}$

Answer 2

$\dfrac{4^n*2^n^-^1}{4^n^+^1} = \dfrac{ (2)^2^{n}* 2^{n-1} }{ (2)^2^{(n+1)} } = \dfrac{ 2^{2n+n-1} }{ 2^{2n+2} } = \dfrac{ 2^{3n-1} }{ 2^{2n+2} } = 2^{(3n-1)-(2n+2)} = \\ \\ \\ 2^{3n-1-2n-2} = \boxed{ 2^{n-3}= \frac{2^n}{8} }$