Medium MCQ +4 / -1 PYQ · JEE Mains 2020

The product $${2^{{1 \over 4}}}{.4^{{1 \over {16}}}}{.8^{{1 \over {48}}}}{.16^{{1 \over {128}}}}$$ ... to $\infty$ is equal to :

  1. A ${2^{{1 \over 4}}}$
  2. B ${2^{{1 \over 2}}}$ Correct answer
  3. C 1
  4. D 2

Solution

$${2^{{1 \over 4}}}{.4^{{1 \over {16}}}}{.8^{{1 \over {48}}}}{.16^{{1 \over {128}}}}$$ ... <br><br>= ${2^{{1 \over 4} + {2 \over {16}} + {3 \over {48}} + ...\infty }}$ <br><br>= ${2^{{1 \over 4} + {1 \over 8} + {1 \over {16}} + ...\infty }}$ <br><br>= ${2^{\left( {{{{1 \over 4}} \over {1 - {1 \over 2}}}} \right)}}$ <br><br>= ${2^{{1 \over 2}}}$

About this question

Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression

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