Solving for x in a series equation
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.
$\endgroup$ 33 Answers
$\begingroup$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*}
$\endgroup$ 2 $\begingroup$$\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} $
$\endgroup$ $\begingroup$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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