Write the number $\frac{2\mathrm{.}1}{b}ar(13)=2\mathrm{.}1131313$dots $as$ a ratio $of$ integers.

Solution

$2\mathrm{.}1131313$dots$=2\mathrm{.}1+\frac{13}{{10}^3}+\frac{13}{{10}^5}+\frac{13}{{10}^7}+$cdots

After the first term $we$ have a geometric series with $a=\frac{13}{{10}^3}$ and $r=$

Therefore $we$ have the following.

$\frac{2\mathrm{.}1}{b}ar(13)=2\mathrm{.}1+\frac{\frac{13}{{10}^3}}{1-}$

$=2\mathrm{.}1+\frac{\frac{13}{1000}}{\frac{99}{100}}$

$=21,+\frac{990}{?}$

$=$