<p>If \(\left(\dfrac{8}{x^2} + \dfrac{6}{x} + 4\right)^{10} = \sum_{r=0}^{20} a_r \left(\dfrac{2}{x}\right)^r\), then find the value of \(\dfrac{a_7}{a_{13}}\).</p>
Step-by-Step Solution
Key Concept: Rewrite the trinomial as a perfect square in terms of (2/x) to identify the coefficient pattern, then use the binomial expansion structure to find the ratio of coefficients without computing them explicitly.
<p><strong>Step 1:</strong> Recognize the perfect square pattern in the trinomial.</p><p>Notice that: $\dfrac{8}{x^2} + \dfrac{6}{x} + 4 = \left(\dfrac{2}{x}\right)^2 + 2 \cdot \dfrac{2}{x} \cdot 2 + 2^2 = \left(\dfrac{2}{x} + 2\right)^2$</p><p><strong>Step 2:</strong> Rewrite the given expression using this factorization.</p><p>$\left(\dfrac{8}{x^2} + \dfrac{6}{x} + 4\right)^{10} = \left[\left(\dfrac{2}{x} + 2\right)^2\right]^{10} = \left(\dfrac{2}{x} + 2\right)^{20}$</p><p><strong>Step 3:</strong> Apply the binomial theorem.</p><p>$\left(\dfrac{2}{x} + 2\right)^{20} = \sum_{k=0}^{20} \binom{20}{k} \left(\dfrac{2}{x}\right)^k (2)^{20-k}$</p><p><strong>Step 4:</strong> Identify the coefficients.</p><p>Comparing with $\sum_{r=0}^{20} a_r \left(\dfrac{2}{x}\right)^r$, we have:</p><p>$a_r = \binom{20}{r} 2^{20-r}$</p><p><strong>Step 5:</strong> Calculate the ratio.</p><p>$\dfrac{a_7}{a_{13}} = \dfrac{\binom{20}{7} 2^{13}}{\binom{20}{13} 2^{7}} = \dfrac{\binom{20}{7}}{\binom{20}{13}} \cdot 2^6$</p><p>Since $\binom{20}{7} = \binom{20}{13}$ (complementary binomial coefficients):</p><p>$\dfrac{a_7}{a_{13}} = 1 \cdot 2^6 = 64$</p><p>∴ Answer: 64</p>
Correct Answer: 8