Showing that a sequence containing the floor function follows an equality

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Show that $\Big \lfloor \frac{n+1}{ 2 } \Big \rfloor + \Big \lfloor \frac{n+2}{ 2^2 } \Big \rfloor + \Big \lfloor \frac{n+2^2}{ 2^3 } \Big \rfloor + \dotsb + \Big \lfloor \frac{n+2^k}{ 2^{k+1} } \Big \rfloor +\dotsb = n$, for every positive integer n. Note that this is not an infinite sum since $0<\dfrac{n+2^k}{ 2^{k+1} }<1$ for sufficiently large k, which makes all terms from there equal 0.

This is my attempt:

Consider the first 3 terms:

$$\left\lfloor \bigg(\frac{n}{2}+\frac12\bigg)\right\rfloor, \ \left\lfloor\bigg(\frac{n}{4}+\frac12\bigg)\right\rfloor, \ \left\lfloor\bigg(\frac{n}{8}+\frac12\bigg)\right\rfloor.$$

Hence, we can represent the series as the finite sum:

$$\sum_{k=1}^\infty \left\lfloor\frac{n+2^{k-1}}{2^k}\right\rfloor$$

Consider the first 4 cases:

$n=1$

$$\sum_{k=0}^\infty \left\lfloor\frac{1+2^{k-1}}{2^k}\right\rfloor=\sum_{k=0}^\infty\left\lfloor\frac{1}{2^k}\right\rfloor+\left\lfloor\frac{2^{k-1}}{2^k}\right\rfloor=\sum_{k=0}^\infty\left\lfloor\frac{1}{2^{k}}\right\rfloor+\left\lfloor\frac{1}{2}\right\rfloor$$

$n=2$

$$\sum_{k=0}^\infty \left\lfloor\frac{2+2^{k-1}}{2^k}\right\rfloor=\sum_{k=0}^\infty\left\lfloor\frac{2}{2^k}\right\rfloor+\left\lfloor\frac{2^{k-1}}{2^k}\right\rfloor=\sum_{k=0}^\infty\left\lfloor\frac{1}{2^{k-1}}\right\rfloor+\left\lfloor\frac{1}{2}\right\rfloor$$

$n=3$

$$\sum_{k=0}^\infty \left\lfloor\frac{3+2^{k-1}}{2^k}\right\rfloor=\sum_{k=0}^\infty\left\lfloor\frac{3}{2^k}\right\rfloor+\left\lfloor\frac{2^{k-1}}{2^k}\right\rfloor=\sum_{k=0}^\infty\left\lfloor\frac{3}{2^k}\right\rfloor+\left\lfloor\frac{1}{2}\right\rfloor$$

$n=4$

$$\sum_{k=0}^\infty \left\lfloor\frac{4+2^{k-1}}{2^k}\right\rfloor=\sum_{k=0}^\infty\left\lfloor\frac{4}{2^k}\right\rfloor+\left\lfloor\frac{2^{k-1}}{2^k}\right\rfloor=\sum_{k=0}^\infty\left\lfloor\frac{2}{2^{k-2}}\right\rfloor+\left\lfloor\frac{1}{2}\right\rfloor$$

For the terms $\frac{1}{2^k},\cdots,\frac{2}{2^{k-2}}$ we obtain 0, $\forall\ k\geq1$ and since $\left\lfloor \frac{1}{2}\right\rfloor=0$, we obtain all terms equal to zero, $\forall\ k\geq1$.

But this is wrong.

Any ideas how to solve this?

Thanks