<p>Sum the series: \(1 + \dfrac{1}{3} + \dfrac{1 \cdot 3}{3 \cdot 6} + \dfrac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9} + \ldots\) to \(\infty\).</p>
Step-by-Step Solution
Key Concept: Recognize the general term as coefficients from the binomial expansion of (1+x)^(-1/2), then use the binomial series formula to sum it. The numerators follow the double factorial pattern (2n-1)!! and denominators are 3^n·n!, which appear in (1-4/3)^(-1/2).
<p><strong>Step 1: Identify the general term</strong></p><p>The series is: 1 + 1/3 + (1·3)/(3·6) + (1·3·5)/(3·6·9) + ...</p><p>General term: t_n = [1·3·5·...·(2n-1)] / [3·6·9·...·(3n)]</p><p>Rewrite: t_n = [1·3·5·...·(2n-1)] / [3^n · (1·2·3·...·n)] = [(2n-1)!!] / [3^n · n!]</p><p><strong>Step 2: Express using binomial coefficients</strong></p><p>Note that (2n-1)!! = (2n)! / [2^n · n!], so:</p><p>t_n = (2n)! / [2^n · n! · 3^n · n!] = (1/2^n) · C(2n,n) · (1/3)^n</p><p>This matches the binomial series coefficient from (1+x)^(-1/2) with x = -4/3.</p><p><strong>Step 3: Apply binomial series</strong></p><p>From (1+x)^(-1/2) = Σ C(-1/2, n)·x^n, we have:</p><p>Σ [1·3·5·...·(2n-1)] / [2^n · n!] · x^n = (1+x)^(-1/2)</p><p>With x = -4/3:</p><p>S = (1 - 4/3)^(-1/2) = (-1/3)^(-1/2) = √3</p><p><strong>∴ Answer: √3</strong></p>
Correct Answer: √3