10.3.54 repeating decimal + geometric series

[karush]

$\tiny{206.10.3.54}$
$\text{Write the repeating decimal first as a geometric series} \\$
$\text{and then as fraction (a ratio of two intergers)} \\$
$\text{Write the repeating decimal as a geometric series} $
$6.94\overline{32}=6.94323232 \\$
$\displaystyle A.\ \ \ 6.94\overline{32}=\sum_{k=0}^{\infty}6.94(0.1)^k \\$
$\displaystyle B.\ \ \ 6.94\overline{32}=0.0032+\sum_{k=0}^{\infty}6.94(0.001)^k \\$
$\displaystyle C.\ \ \ 6.94\overline{32}=6.94+\sum_{k=0}^{\infty}0.0032(0.01)^k$

chose c but guessed?

 

[MarkFL]

\(\displaystyle 6.94\overline{32}=\frac{694}{100}+\frac{32}{100}\cdot\frac{1}{99}=\frac{694}{100}+\frac{32}{10000}\cdot\frac{1}{1-\dfrac{1}{100}}=\frac{694}{100}+\frac{32}{100}\sum_{k=1}^{\infty}\left(\left(\frac{1}{100}\right)^k\right)=\frac{694}{100}+\frac{32}{10000}\sum_{k=0}^{\infty}\left(\left(\frac{1}{100}\right)^k\right)\)

This is equivalent to choice c).

\(\displaystyle 6.94\overline{32}=\frac{694}{100}+\frac{32}{9900}=\frac{34369}{4950}\)

 

[karush]

thanks couldn't find any example on how to do this one.
kinda strange prob!em