So I'm a bit late to the party but it seems to me that the other answers, while correct, are incomplete. Reading between the lines it seems like you're trying to evaluate the series
$$\sum_{k = 0}^{\infty} x^{\lfloor \frac{k}{a-b} \rfloor},\quad\text{integers }a,b \text{ and } 0 \leqslant x < 1.$$
You'd like to be able to pull out the index variable $k$ and then treat it like a regular geometric progression. (Note that $\lfloor k \rfloor = k$ for all integers $k$.) However, like others have stated, this is invalid. You can sometimes do it, but not here. The general rule (for real $r$) is
$$\lfloor nr \rfloor = n \lfloor r \rfloor \quad \iff \quad \{r\} < \tfrac 1 n,\quad \text{integer } n>0,$$
where $\{r\} := r - \lfloor r \rfloor$ is the fractional part of $r$.
Anyway, let's see why this doesn't work for our summand. First, we'll put $m = a-b$ to clean things up. We cannot have $m = 0$, and the series diverges for negative $m$, so we must have $m > 0$. Consider the terms after the first, those with $k > 0$. When is it true that $\lfloor k/m \rfloor = \lfloor 1/m \rfloor k$? Precisely when $\{1/m\} < 1/k$. If $m=1$ this certaintly holds, but if $m>1$ we have
$$ \{1/m\} < 1/k \quad \iff\quad 1/m < 1/k \quad \iff \quad k < m.$$
So it works for the first few terms, but once $k$ is large enough it's always wrong to pull it out! This makes sense: $\lfloor 1/m \rfloor$ = 0 for $m > 1$, so multiplying it by $k$ on the outside will never get us back up to $\lfloor k/m \rfloor$, unless $\lfloor k/m \rfloor$ is $0$.
In case you're still interested, here's the proper way to do the sum: realize that the first $m$ terms are equal to $x^0$, the next $m$ terms are equal to $x^1$, and so forth. So the sum is equal to
$$m\sum_{j=0}^{\infty} x^j = \frac{m}{1-x}.$$
