Question 706048
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This is the most interesting problem I've seen on this site in a long time.


The first thing to do is to factor *[tex \LARGE x] out of all of the terms, so that just becomes a factor in the answer.


Then you have a sum of radicals.  Note that each of the radicands is one power of 3 larger than the previous.  The pattern becomes evident when you replace the radical with a fractional exponent.


*[tex \LARGE \sqrt{3}\ =\ 3^{\frac{1}{2}}]


*[tex \LARGE \sqrt{9}\ =\ 3^1\ =\ 3]


*[tex \LARGE \sqrt{27}\ =\ 3^{\frac{3}{2}}\ =\ 3\ \cdot\ 3^{\frac{1}{2}}]


and so on...


So you could write the first six terms thus:


*[tex \LARGE \sqrt{3}\ +\ 3\ +\ 3\sqrt{3}\ +\ 9\ +\ 9\sqrt{3}\ +\ 27]


Now if you rearrange the terms so that the rational terms are in one group and the irrational terms are in another group:


*[tex \LARGE \{\sqrt{3}\ +\ 3\sqrt{3}\ +\ 9\sqrt{3}\}\ +\ \{3\ +\ 9\ +\ 27}]


Then factor the radical out of the first grouping (putting the *[tex \LARGE x] back in):


*[tex \LARGE x\left(\sqrt{3}\left\{1\ +\ 3\ +\ 9\right\}\ +\ \left\{3\ +\ 9\ +\ 27\right\}\right)]


The sum becomes trivial


So for 12 terms:


*[tex \LARGE x\left(\sqrt{3}x\left\{1\ +\ 3\ +\ 9\ +\ 27\ +\ 81\ +\ 243\right\}\ +\ \left\{3\ +\ 9\ +\ 27\ +\ 81\ +\ 243\ +\ 729\right\}\right)]


But then we get to make it even simpler because


*[tex \LARGE \sum_{i=0}^n\,r^i\ =\ \frac{1\ -\ r^{n+1}}{1\ -\ r}]


from which we can derive


*[tex \LARGE \sum_{i=1}^n\,r^i\ =\ \frac{1\ -\ r^{n+1}}{1\ -\ r}\ -\ 1]


So your 12 term sum becomes


*[tex \LARGE x\left(\sqrt{3}\left(\frac{1\ -\ 3^6}{1\ -\ 3}\right)\ +\ \left(\frac{1\ -\ 3^7}{1\ -\ 3}\ -\ 1\right)\right)]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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