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