We'll let $M=2^k$ throughout.
Note that $$f(j)=j-M\left\lfloor\frac{j}{M}\right\rfloor$$
is just the modulus operator - it is equal to the smallest positive $n$ such that $j\equiv n\pmod {M}$
So that means $f(0)=0, f(1)=1,...f(M-1)=M-1,$ and $f(j+M)=f(j)$.
This means that we can write:
$$F(z)=\sum_{j=0}^{\infty} f(j)z^{j}= \left(\sum_{j=0}^{M-1} f(j)z^{j}\right)\left(\sum_{i=0}^\infty z^{Mi}\right)$$
But $$\sum_{i=0}^\infty z^{Mi} = \frac{1}{1-z^{M}}$$
and $f(j)=j$ for $j=0,...,2^k-1$, so this simplifies to:
$$F(z)=\frac{1}{1-z^{M}}\sum_{j=0}^{M-1} jz^j$$
Finally, $$\sum_{j=0}^{M-1} jz^j = z\sum_{j=1}^{M-1} jz^{j-1} =z\frac{d}{dz}\frac{z^M-1}{z-1}=\frac{(M-1)z^{M+1}-Mz^{M}+z}{(z-1)^2}$$
So:
$$F(z)=\frac{(M-1)z^{M+1}-Mz^{M}+z}{(z-1)^2(1-z^{M})}$$
Your final sum is: $$\alpha F(1-\alpha)$$ bearing in mind, again, that $M=2^k$
It’s not generally true that $\lfloor 3x\rfloor=3\lfloor x\rfloor$; try $x=\frac13$, for example. One very straightforward approach is to let $n=\lfloor x\rfloor$, so that $n\le x<n+1$, and consider three cases:
- $n\le x<n+\frac13$;
- $n+\frac13\le x<n+\frac23$; and
- $n+\frac23\le x<n+1$.
Best Answer
What you're trying to show is not always true. For example, let $n = 4$, so $n - 1 = 3$, and $k = 1$, so $2^k = 2$. This means $n + 2^k - 1 = 5$. However, using that each summation term is non-negative, the first few terms then become
$$\begin{equation}\begin{aligned} & \sum_{i=1}^{\infty} \left(\left\lfloor{\frac{n+2^k-1}{2^i}}\right\rfloor - \left\lfloor{\frac{n-1}{2^i}}\right\rfloor\right) \\ & = \left(\left\lfloor\frac{5}{2}\right\rfloor - \left\lfloor\frac{3}{2}\right\rfloor\right) + \left(\left\lfloor\frac{5}{4}\right\rfloor - \left\lfloor\frac{3}{4}\right\rfloor\right) + \ldots \\ & = (2 - 1) + (1 - 0) + \ldots \\ & = 1 + 1 + \ldots \\ & = 2 \end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Note I've used that all of the remaining terms, i.e., the "$\ldots$" part, are $0$. This is larger than your right side of $2^{k} - 1 = 1$.
Note your statement would be correct with an appropriate restriction between $k$ and $n$. In particular, something like $2^k \gt n - 1$ works. This is because, for $i \le k$, each of the summation terms becomes
$$\begin{equation}\begin{aligned} \left\lfloor{\frac{n+2^k-1}{2^i}}\right\rfloor - \left\lfloor{\frac{n-1}{2^i}}\right\rfloor & = \left(\left\lfloor{\frac{2^k}{2^i} + \frac{n-1}{2^i}}\right\rfloor\right) - \left\lfloor{\frac{n-1}{2^i}}\right\rfloor \\ & = \left(2^{k-i} + \left\lfloor{\frac{n-1}{2^i}}\right\rfloor\right) - \left\lfloor{\frac{n-1}{2^i}}\right\rfloor \\ & = 2^{k-i} \end{aligned}\end{equation}\tag{2}\label{eq2A}$$
For $i \gt k$, then $2^i \ge 2(2^k) \gt n + 2^k - 1$ and $2^i \gt n - 1$, so both floor terms become $0$. Thus, the total sum would be that of a geometric series of
$$\sum_{i=1}^{k}(2^{k-i}) = 2^k - 1 \tag{3}\label{eq3A}$$
which then matches your right side.