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Question Number 167955 by peter frank last updated on 30/Mar/22

Answered by JDamian last updated on 30/Mar/22

$$6^{\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\cdot \cdot \cdot }=6^1=6$$

Answered by alephzero last updated on 30/Mar/22

$$6^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...}={\exp }_6\left(\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{2^n}\right)$$$$S=\frac{a_1}{1-q}{a}_1=\frac{1}{2}\wedge q=\frac{1}{2}$$$$\Rightarrow S=\frac{\left(\frac{1}{2}\right)}{(1-\frac{1}{2})}=1$$$$\Rightarrow 6^{\frac{1}{2}}{6}^{\frac{1}{4}}{6}^{\frac{1}{8}}...=6^1=6◼QED$$

Commented by peter frank last updated on 30/Mar/22

$$\mathrm{thank}\mathrm{you}$$

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