\( \frac { 5 ^ { n - 1 } \cdot 2 ^ { n + 2 } } { 10 ^ { n } - 10 ^ { n - 1 } \cdot 2 } \)
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To simplify the expression \( \frac { 5 ^ { n - 1 } \cdot 2 ^ { n + 2 } } { 10 ^ { n } - 10 ^ { n - 1 } \cdot 2 } \), let's start with the denominator: First, rewrite \( 10^n \) as \( 10^n = 2^n \cdot 5^n \). This means: \[ 10^{n-1} \cdot 2 = 2 \cdot 10^{n-1} = 2 \cdot 2^{n-1} \cdot 5^{n-1} = 2^n \cdot 5^{n-1} \] Now, substituting back into the denominator gives us: \[ 10^n - 10^{n-1} \cdot 2 = 10^n - 10^{n-1} \cdot 2^n \cdot 5^{n-1} = 2^n \cdot 5^n - 2^n \cdot 5^{n-1} = 2^n \cdot (5^n - 5^{n-1}) = 2^n \cdot 5^{n-1} \cdot (5 - 1) = 2^n \cdot 5^{n-1} \cdot 4 \] So we can now rewrite the entire expression: \[ \frac{5^{n-1} \cdot 2^{n+2}}{2^n \cdot 5^{n-1} \cdot 4} \] Next, we can cancel out \( 5^{n-1} \) in the numerator and the denominator: \[ \frac{2^{n+2}}{2^n \cdot 4} \] Now we simplify \( \frac{2^{n+2}}{2^n \cdot 4} = \frac{2^{n+2}}{2^n \cdot 2^2} = \frac{2^{n+2}}{2^{n+2}} = 1 \] Thus, the expression simplifies to: \[ \boxed{1} \]