$$ \frac { 1 } { r } imes \frac { 1 - ( \frac { 1 } { r } ) ^ { n } } { 1 - \frac { 1 } { r } } \geq \frac { 9 r } { 100 } \Rightarrow 1 - \frac { 1 } { r ^ { n } } \geq \frac { 9 r } { 100 } $$

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$$ \frac { 1 } { r } 	imes \frac { 1 - ( \frac { 1 } { r } ) ^ { n } } { 1 - \frac { 1 } { r } } \geq \frac { 9 r } { 100 } \Rightarrow 1 - \frac { 1 } { r ^ { n } } \geq \frac { 9 r } { 100 } $$

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Let's examine the inequality you've provided and the subsequent implication to determine its validity. ### Original Inequality: $$ \frac{1}{r} \times \frac{1 - \left(\frac{1}{r}\right)^n}{1 - \frac{1}{r}} \geq \frac{9r}{100} $$ ### Simplifying the Left-Hand Side (LHS): 1. **Simplify the Fraction:** $$ \frac{1 - \left(\frac{1}{r}\right)^n}{1 - \frac{1}{r}} $$ Multiply numerator and denominator by $$ r $$ to eliminate the fractions: $$ \frac{r \left(1 - \left(\frac{1}{r}\right)^n\right)}{r \left(1 - \frac{1}{r}\right)} = \frac{r - \frac{1}{r^{n-1}}}{r - 1} $$ This simplifies to: $$ \frac{1 - \frac{1}{r^n}}{r - 1} $$ 2. **Multiply by $$ \frac{1}{r} $$:** $$ \frac{1}{r} \times \frac{1 - \frac{1}{r^n}}{r - 1} = \frac{1 - \frac{1}{r^n}}{r(r - 1)} = \frac{1 - \frac{1}{r^n}}{r^2 - r} $$ So, the original inequality becomes: $$ \frac{1 - \frac{1}{r^n}}{r^2 - r} \geq \frac{9r}{100} $$ ### Proposed Implication: $$ 1 - \frac{1}{r^n} \geq \frac{9r}{100} $$ ### Analyzing the Implication: To determine whether the original inequality implies the proposed one, let's explore the relationship between them. 1. **From Original to Proposed:** Starting with: $$ \frac{1 - \frac{1}{r^n}}{r^2 - r} \geq \frac{9r}{100} $$ Multiply both sides by $$ r^2 - r $$ (assuming $$ r > 1 $$ so that $$ r^2 - r > 0 $$): $$ 1 - \frac{1}{r^n} \geq \frac{9r(r^2 - r)}{100} $$ This simplifies to: $$ 1 - \frac{1}{r^n} \geq \frac{9r^3 - 9r^2}{100} $$ **Notice:** The right-hand side is **not** the same as $$ \frac{9r}{100} $$. Instead, it's a cubic expression in terms of $$ r $$. 2. **Conclusion:** The implication you've stated: $$ \frac{1}{r} \times \frac{1 - \left(\frac{1}{r}\right)^n}{1 - \frac{1}{r}} \geq \frac{9r}{100} \Rightarrow 1 - \frac{1}{r^n} \geq \frac{9r}{100} $$ **is not generally valid** because: - The original inequality, after simplification, leads to: $$ 1 - \frac{1}{r^n} \geq \frac{9r^3 - 9r^2}{100} $$ not $$ \frac{9r}{100} $$. - Therefore, unless $$ \frac{9r^3 - 9r^2}{100} \geq \frac{9r}{100} $$, which simplifies to $$ r^3 - r^2 \geq r $$ or $$ r^3 - r^2 - r \geq 0 $$, the implication does not hold. 3. **When Does the Implication Hold?** Let's solve $$ r^3 - r^2 - r \geq 0 $$: $$ r(r^2 - r - 1) \geq 0 $$ - **Roots of $$ r^2 - r - 1 = 0 $$:** $$ r = \frac{1 \pm \sqrt{5}}{2} $$ Approximately: $$ r \approx 1.618 \quad \text{and} \quad r \approx -0.618 $$ - **Analysis:** - For $$ r > 1.618 $$, $$ r(r^2 - r - 1) > 0 $$, so the implication holds. - For $$ 1 < r < 1.618 $$, $$ r(r^2 - r - 1) < 0 $$, so the implication **does not hold**. - For $$ 0 < r < 1 $$, the behavior depends on the specific value of $$ r $$, but generally, the implication does **not** hold. ### Final Remarks: - **Incorrect Implication:** The step from the original inequality to $$ 1 - \frac{1}{r^n} \geq \frac{9r}{100} $$ is only valid under specific conditions (primarily $$ r \geq 1.618 $$), and it's not generally true for all $$ r > 1 $$. - **Recommendation:** Ensure that the conditions under which the implication holds are clearly stated. If you're working within a context where $$ r $$ is guaranteed to be greater than approximately 1.618, then the implication is valid. Otherwise, additional constraints or adjustments are necessary. If you have a specific range for $$ r $$ or additional context, please provide that for a more tailored explanation. The implication $$ 1 - \frac{1}{r^n} \geq \frac{9r}{100} $$ is not generally valid for all $$ r > 1 $$. It holds true only when $$ r \geq 1.618 $$.

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