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Solve for x:

$$ \sum_{n = 1}^{\infty} 7x^{5n} = 36 $$

Just curious as to how you would go about solving a problem like this? It looks as though differential equations come into play, and I'm assuming standard algebra rules are ruled out, where x has to be on one side of the equation. Anyway's, the fact that there's a variable x in the equation throws me into a rut because of the existence of n as well. Any kind of help on how to go about solving this problem would be greatly appreciated.

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  • $\begingroup$ Use the formula for the sum of a geometric progression. $\endgroup$ – Nicholas Oct 29 '15 at 6:16
  • $\begingroup$ $\sum_{i=1}^{\infty}7x^{5n} = 36 \implies \sum_{i=1}^{\infty}x^{5n} = 36/7 => y = x^5; \sum_{i=1}^{\infty}y^{n} = 36/7 \implies $ ????? $\endgroup$ – fleablood Oct 29 '15 at 6:21
  • $\begingroup$ Read up on Geometric Series: mathworld.wolfram.com/GeometricSeries.html $\endgroup$ – fleablood Oct 29 '15 at 6:23
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An infinite geometric series with $|r| < 1$ converges as follows:

$$ \sum_{k = m}^{\infty} ar^k = \frac{ar^m}{1 - r} $$

In your case, $r = x^5$, $a = 7$ and $m = 1$, so we can deduce the following:

\begin{align*} \frac{7x^5}{1 - x^5} &= 36 \\ 7x^5 &= 36(1 - x^5) \\ x &= \left(\frac{36}{43}\right)^{1/5} \end{align*}

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  • $\begingroup$ Excellent, so r stands as the common ratio, correct? And a is our constant in the equation. Just making sure. $\endgroup$ – etree Oct 29 '15 at 6:50
  • $\begingroup$ Yep, that's correct. $\endgroup$ – Gaussian0617 Oct 29 '15 at 6:52
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$\sum _{n=1}^\infty 7x^{5n} = \sum _{n=1}^\infty 7(x^{5})^n= \frac{7}{1-x^5} -7 $

$\frac{7}{1-x^5} -7 =36$

$x^5 = \frac{36}{43} $

$x = (\frac{36}{43} )^{1/5} $

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The sum converges iff $|x^5| < 1$, in which we have the closed form formula $\sum_{n=1}^\infty 7 x^{5n} = 7 \left( \frac{1}{1 - x^5} -1 \right)$, which then allows one to solve for $x$, yielding $x = \left (\frac{36}{43} \right)^{1/5}$.

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