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\( \frac { 5 ^ { n - 1 } \cdot 2 ^ { n + 2 } } { 10 ^ { n } - 10 ^ { n - 1 } \cdot 2 } \)

Ask by Vaughan King. in South Africa
Jan 23,2025

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Tutor-Verified Answer

Answer

The expression simplifies to 1 for all positive integers \( n \).

Solution

Let's simplify the expression: \[ \frac{5^{n - 1} \cdot 2^{n + 2}}{10^n - 10^{n - 1} \cdot 2} \] ### Step 1: Express Everything in Terms of Primes First, note that \(10 = 2 \times 5\). We'll rewrite the denominator using this: \[ 10^n = (2 \times 5)^n = 2^n \cdot 5^n \] \[ 10^{n-1} = (2 \times 5)^{n-1} = 2^{n-1} \cdot 5^{n-1} \] So, the denominator becomes: \[ 10^n - 10^{n-1} \cdot 2 = 2^n \cdot 5^n - 2^{n-1} \cdot 5^{n-1} \cdot 2 \] Simplify the second term in the denominator: \[ 2^{n-1} \cdot 5^{n-1} \cdot 2 = 2^n \cdot 5^{n-1} \] Now, rewrite the denominator: \[ 2^n \cdot 5^n - 2^n \cdot 5^{n-1} = 2^n \cdot 5^{n-1} \cdot (5 - 1) = 2^n \cdot 5^{n-1} \cdot 4 \] So, the denominator simplifies to: \[ 4 \cdot 2^n \cdot 5^{n-1} \] ### Step 2: Rewrite the Numerator The numerator is: \[ 5^{n - 1} \cdot 2^{n + 2} = 5^{n-1} \cdot 2^n \cdot 2^2 = 5^{n-1} \cdot 2^n \cdot 4 \] ### Step 3: Form the Simplified Fraction Now, the expression becomes: \[ \frac{5^{n-1} \cdot 2^n \cdot 4}{4 \cdot 2^n \cdot 5^{n-1}} \] ### Step 4: Cancel Common Terms Notice that \(5^{n-1}\), \(2^n\), and \(4\) appear in both the numerator and the denominator. Cancel them out: \[ \frac{\cancel{5^{n-1}} \cdot \cancel{2^n} \cdot \cancel{4}}{\cancel{4} \cdot \cancel{2^n} \cdot \cancel{5^{n-1}}} = 1 \] ### Conclusion After simplifying, the expression equals **1** for all positive integers \(n\): \[ \frac{5^{n - 1} \cdot 2^{n + 2}}{10^n - 10^{n - 1} \cdot 2} = 1 \]

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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: \[ \boxed{1} \]

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