<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 recognize the exponents form an infinite geometric series. The product becomes 2 raised to the sum of the exponent series.
<p><strong>Step 1:</strong> Express each term as a power of 2:</p><p>2^(1/4) · 4^(1/8) · 8^(1/16) · ... = 2^(1/4) · (2²)^(1/8) · (2³)^(1/16) · ...</p><p>= 2^(1/4) · 2^(2/8) · 2^(3/16) · ...</p><p>= 2^(1/4) · 2^(1/4) · 2^(3/16) · ...</p><p><strong>Step 2:</strong> Identify the exponent pattern:</p><p>Exponents: 1/4, 2/8, 3/16, 4/32, ... = 1/4, 1/4, 3/16, 1/8, ...</p><p>Rewrite: 1/2², 2/2³, 3/2⁴, 4/2⁵, ... = Σ(n=1 to ∞) n/2^(n+1)</p><p><strong>Step 3:</strong> Calculate the sum S = Σ(n=1 to ∞) n/2^(n+1) = (1/2)Σ(n=1 to ∞) n/2^n</p><p>Using formula Σ(n=1 to ∞) nx^n = x/(1-x)² for |x| < 1, with x = 1/2:</p><p>Σ(n=1 to ∞) n/2^n = (1/2)/(1-1/2)² = (1/2)/(1/4) = 2</p><p>Therefore: S = (1/2) · 2 = 1</p><p><strong>Step 4:</strong> The product equals 2^S = 2^1 = 2</p><p>∴ Answer: B (which equals 2)</p>
Correct Answer: B