JEE Main & Advanced JEE Main Paper Phase-I (Held on 09-1-2020 Morning)

  • question_answer
    The product \[{{2}^{\frac{1}{4}}}.\,{{4}^{\frac{1}{16}}}.\,{{8}^{\frac{1}{48}}}.\,{{16}^{\frac{1}{128}}}.....\] to \[\infty \] is equal to                                                         [JEE MAIN Held on 09-01-2020 Morning]

    A) \[{{2}^{\frac{1}{2}}}\]           

    B) \[{{2}^{\frac{1}{4}}}\]

    C) \[2\]                 

    D) \[1\]

    Correct Answer: A

    Solution :

    [a] \[{{2}^{\frac{1}{4}}}.\,\,{{4}^{\frac{1}{16}}}.\,\,{{8}^{\frac{1}{48}}}......\] \[={{2}^{\frac{1}{4}}}.\,\,{{2}^{\frac{1}{8}}}.\,\,{{2}^{\frac{1}{16}}}........\] \[={{2}^{\left( \frac{1}{4}+\frac{1}{8}+\frac{1}{16}+..... \right)}}\] \[={{2}^{\frac{1/4}{1-1/2}}}\] \[={{2}^{\frac{1}{2}}}=\sqrt{2}\]


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