Prove that: (2^ 14 .4^ 18 .8^ 1 16 .16^ 1 32 )=2

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Prove that: $\Big(2^{\frac14}.4^{\frac18}.8^{\frac{1}{16}}.16^{\frac{1}{32}}\dots\infty\Big)=2$

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Answer

$2^{\frac14}.4^{\frac18}.8^{\frac{1}{16}}.16^{\frac{1}{32}}\dots\infty$ $=2^{\frac14}.4^{\frac18}.8^{\frac{1}{16}}.16^{\frac{1}{32}}\dots\infty$ $=\Big(\frac14+\frac18+\frac{3}{16}+\frac{4}{32}+\ \dots\infty\Big)$ $=2$ $=2^{5}\cdots(1)$ $\text{S}=\frac{1}{4}+\frac28+\frac{3}{16}+\frac{4}{32}+\ \dots\infty$ $\text{S}=\Big(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\dots\infty\Big)2$ $\frac{\text{S}}{2}=\frac14+\frac18+\frac{1}{16}+\frac{1}{32}+\dots\infty$ $=\frac{\frac{1}{4}}{1-\frac12}$ $=\frac{1}{4}\times\frac21$ $\text{S}=\frac12$ $\text{S}=1$ Thus, $2^{\frac14}.4^{\frac18}.8^{\frac{1}{16}}.16^{\frac{1}{32}}\dots=2^1=2$

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