<p>\(1 + \dfrac{1}{3}x + \dfrac{1 \times 4}{3 \times 6}x^2 + \dfrac{1 \times 4 \times 7}{3 \times 6 \times 9}x^3 + \cdots\) is equal to</p>
Step-by-Step Solution
Key Concept: Recognize the coefficient pattern as products of arithmetic sequences and match it to the generalized binomial expansion (1+u)^n where n is fractional. The denominators 3, 6, 9, ... and numerators 1, 4, 7, ... form arithmetic progressions with common difference 3.
<p><strong>Step 1:</strong> Identify the general term pattern.</p><p>Numerator of r-th coefficient: 1·4·7·...·(3r-2) = product of r terms with first term 1, common difference 3</p><p>Denominator of r-th coefficient: 3·6·9·...·3r = 3^r · r!</p><p><strong>Step 2:</strong> Express using the Pochhammer symbol or rising factorial.</p><p>Numerator = (1/3)_r · 3^r where (1/3)_r is the rising factorial</p><p>Coefficient of x^r = [1·4·7·...·(3r-2)]/(3·6·9·...·3r) · x^r = [(1/3)(4/3)(7/3)...(3r-2)/3] · x^r</p><p><strong>Step 3:</strong> Recognize this matches the generalized binomial series.</p><p>The series matches (1+x)^α where the coefficients are C(α,r) = α(α-1)(α-2)...(α-r+1)/r!</p><p>Here α = -1/3, giving C(-1/3, r) which produces exactly this coefficient pattern.</p><p><strong>Step 4:</strong> Apply the binomial theorem.</p><p>(1+x)^(-1/3) = 1 + (-1/3)x + (-1/3)(-4/3)x²/2! + (-1/3)(-4/3)(-7/3)x³/3! + ...</p><p>= 1 + (-1/3)x + (1·4)/(3·6)x² + (1·4·7)/(3·6·9)x³ + ...</p><p>∴ Answer: D → **(1+x)^(-1/3)** or equivalently **1/(1+x)^(1/3)** or **∛(1/(1+x))**</p>
Correct Answer: D