<p>Sum the series to infinite terms: \[1 + \frac{2}{6} + \frac{2\cdot5}{6\cdot12} + \frac{2\cdot5\cdot8}{6\cdot12\cdot18} + \cdots\]</p>
Step-by-Step Solution
Key Concept: Recognize the general term as a product of ratios following the pattern (3n-1)/(6n), then express as a binomial series using the form (1-x)^(-1/3) where the coefficients match our numerators and denominators.
<p><strong>Step 1:</strong> Identify the general term. The nth term (n≥1) is: T_n = (2·5·8···(3n-1))/(6·12·18···(6n))</p><p><strong>Step 2:</strong> Rewrite as T_n = [1·2·5·8···(3n-1)]/[6^n · n!]. Factor: T_n = (1/6^n) · [(3n-1)!/(3^n·n!)·(something)] = coefficient from binomial expansion</p><p><strong>Step 3:</strong> Recognize this matches the binomial series: (1-x)^(-1/3) = Σ C(-1/3, n)(-x)^n where C(-1/3, n) = (-1/3)(-4/3)···((2-3n)/3) / n!</p><p><strong>Step 4:</strong> With x = 1/8, we have (1 - 1/8)^(-1/3) = (7/8)^(-1/3) = (8/7)^(1/3) = ∛(8/7) = 2/∛7</p><p><strong>Step 5:</strong> Rationalize: 2/∛7 · (∛49)/(∛49) = 2∛49/7</p><p>∴ Answer: <strong>2∛49/7 or equivalently 2·7^(2/3)/7 = 2/7^(1/3)</strong></p>
Correct Answer: 2