How do you express a periodic/repeating decimal in expanded notation, especially algebraically?

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Let's just say you've been given an assignment to write a number like 0.435435435... in expanded notation. How would you proceed? And how to generalize something like this with algebra?

Thank you very much.

EDIT

Example of expanded notation of a number is on the right side of the following equation:

$0.435 = \left(4\times{}10^{-1}\right) + \left(3\times{}10^{-2}\right) + \left(5\times{}10^{-3}\right)$

The problem is that as soon as we draw a line above the entire sum of these products to try to denote repetition in algebraic form, the line would be perceived (unlike intended) as a grouping symbol.

I apologize for not clarifying things initially.

Thank you.

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3
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Let $x = \smash[t]{0.\overline{435}}$. Since the pattern repeats every three digits, we can express the number as a geometric sum and find its exact (rational) value that way. Observe: \begin{align} &0.\color{red}{435}\color{green}{435}\color{blue}{435}\dots \\ \hline &0.\color{red}{435} \\ {}+{}\; &0.000\color{green}{435} \\ {}+{}\; &0.000000\color{blue}{435} \\ {}+{}\; &\phantom{0.000000000}\cdots \end{align}

Using the ancient geometric sum formula (where $|r| < 1$),
$$ a + ar + ar^2 + ar^3 + \cdots = \frac{a}{1-r}, $$ we calculate \begin{align} x &= \frac{435}{1000^{\phantom{1}}} + \frac{435}{1000^2} + \frac{435}{1000^3} + \cdots \\[4pt] &= \frac{\frac{435}{1000}}{1 - \frac{1}{1000}} \\[4pt] &= \frac{435}{999} \end{align} It happens that this fraction reduces, so we might want to write $$ x = \frac{145}{333}. $$


This generalizes as you might expect, where the denominator ends up being a string of $n$ consecutive $9$s, i.e. the integer $10^n - 1$, where $n$ is the length of the repeating block.

So, for instance, with a block of length $n=2$, the denominator is $10^2-1 = 99$: $$ 0.\overline{47} = 0.474747\dots = \frac{47}{99} $$

6
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In "expanded notation", $$ .435435435\dots=4\times10^{-1}+3\times10^{-2}+5\times10^{-3}+4\times10^{-4}+3\times10^{-5}+5\times10^{-6}+4\times10^{-7}+3\times10^{-8}+5\times10^{-9}+\cdots $$ I don't know what you mean by "denote repetition in algebraic form". Perhaps you would prefer $$ \sum_{n=0}^{\infty}(4\times10^{-1}+3\times10^{-2}+5\times10^{-3})\times10^{-3n} $$