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Question

# Prove that: (21/4 . 41/8 . 81/16. 161/32 ... ∞) = 2.

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Solution

## $\mathrm{LHS}={2}^{\frac{1}{4}}.{4}^{\frac{2}{8}}.{8}^{\frac{3}{16}}.{16}^{\frac{4}{32}}...\infty \phantom{\rule{0ex}{0ex}}={2}^{\left(\frac{1}{4}+\frac{2}{8}+\frac{3}{16}\frac{3}{16}\frac{4}{32}.\infty \right)}\phantom{\rule{0ex}{0ex}}={2}^{\left(\frac{1}{{2}^{2}}+\frac{2}{{2}^{3}}+\frac{3}{{2}^{4}}+\frac{4}{{2}^{5}}+...\infty \right)}\phantom{\rule{0ex}{0ex}}={2}^{\frac{1}{{2}^{2}}\left\{1+\frac{2}{2}+\frac{3}{{2}^{2}}+\frac{4}{{2}^{3}}...\infty \right\}}\phantom{\rule{0ex}{0ex}}={2}^{\frac{1}{{2}^{2}}\left\{\frac{1}{1-\frac{1}{2}}+\frac{1.\frac{1}{2}}{{\left(1-\frac{1}{2}\right)}^{2}}\right\}}\phantom{\rule{0ex}{0ex}}={2}^{\frac{1}{{2}^{2}}\left\{2+2\right\}}\phantom{\rule{0ex}{0ex}}={2}^{1}\phantom{\rule{0ex}{0ex}}=2=\mathrm{RHS}\phantom{\rule{0ex}{0ex}}$

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