Quickie #1

**Soroban** · December 23rd 2006, 04:28 AM

Evaulate: . $\left(\frac{1\!\cdot\!2\!\cdot\!4 \,+ \,2\!\cdot\!4\!\cdot\!8 \,+ \,3\!\cdot\!6\!\cdot\!12 \,+ \,\cdots}{1\!\cdot\!3\!\cdot\!9 \,+ \,2\!\cdot\!6\!\cdot\!18 \,+ \,3\!\cdot\!9\!\cdot\!27 \,+ \,\cdots}\right)^{\frac{1}{3}}$

**CaptainBlack** · December 23rd 2006, 04:55 AM

Originally Posted by Soroban

Evaulate: . $\left(\frac{1\!\cdot\!2\!\cdot\!4 \,+ \,2\!\cdot\!4\!\cdot\!8 \,+ \,3\!\cdot\!6\!\cdot\!12 \,+ \,\cdots}{1\!\cdot\!3\!\cdot\!9 \,+ \,2\!\cdot\!6\!\cdot\!18 \,+ \,3\!\cdot\!9\!\cdot\!27 \,+ \,\cdots}\right)^{\frac{1}{3}}$

. $\left(\frac{1\!\cdot\!2\!\cdot\!4 \,+ \,2\!\cdot\!4\!\cdot\!8 \,+ \,3\!\cdot\!6\!\cdot\!12 \,+ \,\cdots}{1\!\cdot\!3\!\cdot\!9 \,+ \,2\!\cdot\!6\!\cdot\!18 \,+ \,3\!\cdot\!9\!\cdot\!27 \,+ \,\cdots}\right)^{\frac{1}{3}}$

............. $=\frac{(1\cdot 2\cdot 4)^{1/3}}{(1\cdot 3\cdot 9)^{1/3}}\frac{(1+2+3+...)^{1/3}}{(1+2+3+...)^{1/3}}=\frac{2}{3}$

RonL

**Soroban** · December 23rd 2006, 06:02 AM

Well done, Captain!

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