5) \( \frac{2^{2 n-3} \times\left(2^{n+2}\right)^{3}}{\left(2^{n+1}\right)^{5}} \)

To simplify the expression \( \frac{2^{2 n-3} \times\left(2^{n+2}\right)^{3}}{\left(2^{n+1}\right)^{5}} \), we can first apply the properties of exponents. Starting with the numerator: \[ \left(2^{n+2}\right)^{3} = 2^{3(n+2)} = 2^{3n + 6} \] Now we can rewrite the numerator: \[ 2^{2n - 3} \times 2^{3n + 6} = 2^{(2n - 3) + (3n + 6)} = 2^{5n + 3} \] Now for the denominator: \[ \left(2^{n+1}\right)^{5} = 2^{5(n + 1)} = 2^{5n + 5} \] Now we can rewrite the entire expression: \[ \frac{2^{5n + 3}}{2^{5n + 5}} = 2^{(5n + 3) - (5n + 5)} = 2^{3 - 5} = 2^{-2} \] Finally, we know that: \[ 2^{-2} = \frac{1}{2^2} = \frac{1}{4} \] So the simplified expression is: \[ \frac{1}{4} \]