<p>The value of \(2^{1/4} \cdot 4^{1/8} \cdot 8^{1/16} \cdots \infty\) is</p>
Step-by-Step Solution
Key Concept: Express each term as a power of 2, then sum the infinite series of exponents using the standard formula for \(\sum nx^n\).
<p><strong>Step 1:</strong> Rewrite the product: \(2^{1/4} \cdot (2^2)^{1/8} \cdot (2^3)^{1/16} \cdots = 2^{1/4} \cdot 2^{2/8} \cdot 2^{3/16} \cdots\)</p><p><strong>Step 2:</strong> Simplify exponents: \(2^{1/4 + 1/4 + 3/16 + \cdots}\)</p><p><strong>Step 3:</strong> The exponent is: \(\frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \cdots = \sum_{n=1}^{\infty} \frac{n}{2^{n+1}}\)</p><p><strong>Step 4:</strong> Using the formula \(\sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2}\) with \(x = 1/2\): \(\sum_{n=1}^{\infty} \frac{n}{2^{n+1}} = \frac{1}{2} \cdot \frac{1/2}{(1-1/2)^2} = \frac{1}{2} \cdot 2 = 1\)</p><p><strong>Step 5:</strong> Therefore, the product equals \(2^1 = 2\)</p><p>∴ Answer is B.</p>
Correct Answer: B