<p>Sum the series: \[1 + \frac{3}{2^1} + \frac{1\cdot3}{1\cdot2}\cdot\frac{3^2}{2^6} + \frac{1\cdot3\cdot5}{1\cdot2\cdot3}\cdot\frac{3^3}{2^7} + \cdots \text{ to } \infty.\]</p>
Step-by-Step Solution
Key Concept: Recognize that the general term matches the binomial expansion of (1+x)^n where n is a non-integer, specifically (1+x)^(-1/2). The coefficients 1·3·5·...·(2r-1) appear in the expansion of (1+x)^(-1/2).
<p><strong>Step 1:</strong> Identify the general term. The rth term (r ≥ 0) is:</p><p>T_r = (1·3·5·...·(2r-1))/(1·2·3·...·r) · (3^r)/(2^(r+1))</p><p>This equals the binomial coefficient C(-1/2, r) · (3/4)^r where C(n,r) = n(n-1)(n-2)...(n-r+1)/r!</p><p><strong>Step 2:</strong> Recognize this as the binomial series expansion of (1+x)^(-1/2) with x = 3/4:</p><p>(1+x)^(-1/2) = Σ C(-1/2, r) · x^r</p><p><strong>Step 3:</strong> Substitute x = 3/4:</p><p>S = (1 + 3/4)^(-1/2) = (7/4)^(-1/2) = √(4/7) = 2/√7</p><p><strong>Step 4:</strong> Rationalize:</p><p>S = 2√7/7</p><p>∴ Answer: <strong>2√7/7</strong> or <strong>2/√7</strong></p>
Correct Answer: 2