JEE Mains · Maths · STD 11 - 8. sequence and series
The product \(2^{\frac{1}{4}} \cdot 4^{\frac{1}{16}} \cdot 8^{\frac{1}{48}} \cdot 16^{\frac{1}{128}} \cdot \ldots .\) to \(\infty\) is equal to
- A \(2^{\frac{1}{2}}\)
- B \(2^{\frac{1}{4}}\)
- C \(2\)
- D \(1\)
Answer & Solution
Correct Answer
(A) \(2^{\frac{1}{2}}\)
Step-by-step Solution
Detailed explanation
\(2^{\frac{1}{4}} \cdot 4^{\frac{1}{16}} \cdot 8^{\frac{1}{48}} \cdot 16^{\frac{1}{128}} \cdot \ldots . \infty\) \(=2^{\frac{1}{4}} \cdot 2^{\frac{2}{16}} \cdot 2^{\frac{3}{48}} \cdot 2^{\frac{4}{128}} \cdot \ldots \infty\)…
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