<p>Value of \(\left(1 + \dfrac{1}{3}\right)\left(1 + \dfrac{1}{3^2}\right)\left(1 + \dfrac{1}{3^4}\right)\left(1 + \dfrac{1}{3^8}\right) \cdots \infty\) is equal to</p>
Step-by-Step Solution
Key Concept: Recognize this as a telescoping infinite product using the algebraic identity (1-x)(1+x)(1+x²)(1+x⁴)... = 1/(1-x). Multiply and divide by (1 - 1/3) to collapse the product into a simple fraction.
<p><strong>Step 1:</strong> Denote the product as P = (1 + 1/3)(1 + 1/3²)(1 + 1/3⁴)(1 + 1/3⁸)⋯</p><p><strong>Step 2:</strong> Multiply and divide by (1 - 1/3):</p><p>P · (1 - 1/3) = (1 - 1/3)(1 + 1/3)(1 + 1/3²)(1 + 1/3⁴)⋯</p><p><strong>Step 3:</strong> Apply difference of squares repeatedly: (1 - 1/3)(1 + 1/3) = 1 - 1/3² = (1 - 1/3²)(1 + 1/3²) = 1 - 1/3⁴, and so on.</p><p><strong>Step 4:</strong> The telescoping product gives: P · (1 - 1/3) = 1 - 1/3^∞ = 1 - 0 = 1</p><p><strong>Step 5:</strong> Therefore: P = 1/(1 - 1/3) = 1/(2/3) = <strong>3/2</strong></p><p>∴ Answer: C (which equals <strong>3/2</strong>)</p>
Correct Answer: C