<p>If \(\left(2+\dfrac{x}{3}\right)^{55}\) is expanded in the ascending powers of \(x\) and the coefficients of powers of \(x\) in two consecutive terms of the expansion are equal, then these terms are</p>
Step-by-Step Solution
Key Concept: In binomial expansion, if coefficients of consecutive terms are equal, then C(n,r) = C(n,r+1), which only occurs when r and r+1 are equidistant from n/2. Use the equality condition to find which specific terms have equal coefficients.
<p><strong>Step 1:</strong> General term in expansion of (2 + x/3)^55 is:</p><p>T_(r+1) = C(55,r) · 2^(55-r) · (x/3)^r = C(55,r) · 2^(55-r) · (1/3^r) · x^r</p><p><strong>Step 2:</strong> Coefficient of x^r is: C(55,r) · 2^(55-r) / 3^r</p><p><strong>Step 3:</strong> For consecutive terms r and (r+1), if coefficients are equal:</p><p>C(55,r) · 2^(55-r) / 3^r = C(55,r+1) · 2^(55-r-1) / 3^(r+1)</p><p><strong>Step 4:</strong> Simplify:</p><p>C(55,r) · 2 / 3 = C(55,r+1)</p><p>C(55,r) / C(55,r+1) = 3/2</p><p><strong>Step 5:</strong> Using C(n,r)/C(n,r+1) = (r+1)/(n-r):</p><p>(r+1)/(55-r) = 3/2</p><p>2(r+1) = 3(55-r)</p><p>2r + 2 = 165 - 3r</p><p>5r = 163</p><p>r = 32.6</p><p><strong>Step 6:</strong> Since r must be integer, we check r = 32 and r = 33. The two consecutive terms are <strong>T_33 and T_34</strong> (the 33rd and 34th terms when counting from the first term).</p><p>∴ Answer: D</p>
Correct Answer: D