jmc

Recall that a terminating decimal can be written as $\frac{a}{10^b}=\frac{a}{2^b\cdot 5^b}$ where $a$ and $b$ are integers. Since $k$ can be expressed as a terminating decimal, then $1+2x=5^b$, as $1+2x$ is odd for all $x$ and, thus, can't equal $2^b$ or $10^b$. Thus, our sum is equal to $\frac{1}{5}+\frac{1}{25}+\frac{1}{125}+\cdots =\frac{\frac{1}{5}}{1-\frac{1}{5}}=\boxed{\frac{1}{4}}$, by the formula $a/(1-r)$ for the sum of an infinite geometric series with common ratio $r$ (between $-1$ and 1) and first term $a$.