<p>Find the third term in the expansion of \((3+2x)^{3/5}\) if \(|x| > \frac{3}{2}\).</p>
Step-by-Step Solution
Key Concept: For binomial expansion of (a+b)^n where n is not a positive integer, use the generalized binomial theorem: (a+b)^n = a^n[1 + (b/a)]^n, then expand (1+y)^n = 1 + ny + n(n-1)y²/2! + ... The third term corresponds to the term with y², and we must identify it correctly by rewriting in standard form.
<p><strong>Step 1: Identify the correct form for convergence.</strong></p><p>Since |x| > 3/2, we rewrite: (3+2x)^(3/5) = (2x)^(3/5)[1 + 3/(2x)]^(3/5)</p><p><strong>Step 2: Expand (1+y)^(3/5) where y = 3/(2x).</strong></p><p>Using generalized binomial: (1+y)^(3/5) = 1 + (3/5)y + [(3/5)(3/5 - 1)/(2!)]y² + ...</p><p>= 1 + (3/5)y + [(3/5)(-2/5)/2]y² + ...</p><p>= 1 + (3/5)y - (3/25)y² + ...</p><p><strong>Step 3: Substitute y = 3/(2x).</strong></p><p>The expansion becomes: (2x)^(3/5)[1 + (3/5)·(3/2x) - (3/25)·(9/4x²) + ...]</p><p>= (2x)^(3/5)[1 + 9/(10x) - 27/(100x²) + ...]</p><p><strong>Step 4: Identify the third term.</strong></p><p>After multiplying by (2x)^(3/5), the third term (in descending powers of x) is:</p><p>(2x)^(3/5) · [-27/(100x²)] = -(27/100)·2^(3/5)·x^(3/5 - 2)</p><p>= <strong>-(27/100)·2^(3/5)·x^(-7/5)</strong> or <strong>-27·2^(3/5)/(100x^(7/5))</strong></p><p>∴ Answer: <strong>-27·∛4/(100x^(7/5))</strong> or equivalently <strong>-(27/100)·2^(3/5)·x^(-7/5)</strong></p>
Correct Answer: -27