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I found in this response that

$\frac29 + \frac{2}{9^2}+\frac{2}{9^3}+\cdots=\frac14$ implies that the base-$3$ expansion of $1/4$ is $0.020202...$. I am very new to the concept of base-$3$ expansions. Could someone please explain how this statement is true?

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    $\begingroup$ Compare your expression with $0\cdot \frac{1}{3}+2\cdot \frac{1}{3^2}+0\cdot \frac{1}{3^3}+2\cdot\frac{1}{3^4}+0\cdot \frac{1}{3^5}+2\cdot\frac{1}{3^6}+\dots$ and then look at the definition of base-n expansions. $\endgroup$
    – JMoravitz
    Apr 2 at 3:11
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    $\begingroup$ Build intuition by remembering how base-10 expansions work. $0.51042$ for instance corresponding to $5$ tenths and $1$ hundredth, zero thousandths, four ten-thousandths, and two hundred-thousandths... that is $0.51042 = 5\cdot 10^{-1}+1\cdot 10^{-2}+0\cdot 10^{-3}+4\cdot 10^{-4}+2\cdot 10^{-5}$ $\endgroup$
    – JMoravitz
    Apr 2 at 3:13
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    $\begingroup$ @JMoravitz, this is good, and you should copy it into an answer. $\endgroup$ Apr 2 at 3:14
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    $\begingroup$ It's exactly the same as base $10$. we have $0.020202020..... $ is defined in any base $b$ as $0\cdot \frac 1b + 2\times \frac 1{b^2} + 0\cdot \frac 1{b^3} + 2\times \frac 1{b^4} + ....$. That is the definition of base $b$. So if $b= 3$ the we have $0.02020202... = 2\frac 1{3^2} + 2\frac 1{3^4} + 2\frac 1{3^6} = \frac 2{9} + \frac 2{9^2} + \frac 2{9^3} + .... = \frac 14$. .... Of course the significant part is knowing $ \frac 2{9} + \frac 2{9^2} + \frac 2{9^3} + .... = \frac 14$ $\endgroup$
    – fleablood
    Apr 2 at 3:19
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By definition $0.a_1a_2a_3a_4.....$ in base $b$ is by definition equal to

$\frac {a_1}{b} + \frac {a_2}{b^2} + \frac {a_3}{b^3}+\frac {a_4}{b^4} + .....$

So in base $3$ we will have $0.020202020..... =$

$\frac 03 + \frac 2{3^2} + \frac 0{3^3} + \frac 0{3^4} + .... $.

As only the even spaces have digits and the odd spaces are all $0$ then this is equal to

$\frac 2{3^{2\cdot 1}} + \frac 2{3^{2\cdot 2}} + \frac 2{3^{2\cdot 3}}+ ....=

$\frac 2{9} + \frac 2{9^2} + \frac 2{9^3}+....$

And that's that.

We can use geometric series that $\frac 29 + \frac 2{9^2} + ... = 2(\sum \frac 1{9^k}) = 2\frac {1}{9-1} = \frac 14$.

Alternatively $0.0202020202020......= k$.

So $3^2k = 2.02020202020..... = 2 + k$.

So $9k = 2+k$ and $8k = 2$ and $k = \frac 14$.

This is the exact same thing as in base $10$ that $0.020202020... = \frac 2{100} + \frac 2{100^2} + \frac 2{100^3} +.....= \frac 2{100-1} =\frac 2{99}$.

And that if $0.02020202020.... = m$ then

$100m =2.0202020202.... = 2 + m$

$100m = 2+m$

$99m = 2$ and $m= \frac 1{99}$.

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Converting between base $b$ and base $b^k$ for $k\ge2$ is extremely simple: one base-$b^k$ digit corresponds one-for-one with $k$ base-$b$ digits.

The given result shows that the base-$9$ expansion of $\frac14$ is $0.\overline2$. Each digit $2$ becomes $02$ in base $3$, yielding the desired result $\frac14=0.\overline{02}_3$.

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