<p>In <span>\(\left(\sqrt[3]{2} + \dfrac{1}{\sqrt[3]{3}}\right)^n\)</span>, if the ratio of 7th term from the beginning to the 7th term from the end is <span>\(1/6\)</span>, then find the value of <span>\(n\)</span>.</p>
Step-by-Step Solution
Key Concept: In a binomial expansion, the rth term from the beginning is T_r and the rth term from the end is T_(n-r+2). Setting up the ratio and using the property that T_r depends on binomial coefficients and powers reveals a symmetric relationship that uniquely determines n.
<p><strong>Step 1:</strong> The general term in the expansion of (√[3]{2} + 1/√[3]{3})^n is:</p><p>T_(r+1) = C(n,r) · (√[3]{2})^(n-r) · (1/√[3]{3})^r = C(n,r) · 2^((n-r)/3) · 3^(-r/3)</p><p><strong>Step 2:</strong> The 7th term from beginning is T_7 = T_(6+1), so r = 6:</p><p>T_7 = C(n,6) · 2^((n-6)/3) · 3^(-6/3) = C(n,6) · 2^((n-6)/3) · 3^(-2)</p><p><strong>Step 3:</strong> The 7th term from end is T_(n-7+2) = T_(n-5). This means we need the (n-6)th term (where r = n-6):</p><p>T_(n-5) = C(n,n-6) · 2^((n-(n-6))/3) · 3^(-(n-6)/3) = C(n,6) · 2^(6/3) · 3^(-(n-6)/3)</p><p><strong>Step 4:</strong> Set up the ratio T_7/T_(n-5) = 1/6:</p><p>[C(n,6) · 2^((n-6)/3) · 3^(-2)] / [C(n,6) · 2^2 · 3^(-(n-6)/3)] = 1/6</p><p><strong>Step 5:</strong> Simplifying (C(n,6) cancels):</p><p>2^((n-6)/3 - 2) · 3^(-2 + (n-6)/3) = 1/6</p><p>2^((n-12)/3) · 3^((n-12)/3) = 1/6 = 2^(-1) · 3^(-1)</p><p><strong>Step 6:</strong> This gives us:</p><p>(2 · 3)^((n-12)/3) = (2 · 3)^(-1)</p><p>6^((n-12)/3) = 6^(-1)</p><p><strong>Step 7:</strong> Therefore:</p><p>(n-12)/3 = -1</p><p>n - 12 = -3</p><p>n = 9</p><p>∴ <strong>Answer: n = 9</strong></p>
Correct Answer: n = 9