Prove that
$$\lfloor{x}\rfloor=\bigg\lfloor{\frac{x+1}{2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2}{2^2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2^2}{2^3}}\bigg\rfloor+ \ldots$$ $x\geq{0}$
How should I start? Any hints would be helpful.
Also how would the expression change for $x\lt{0}.$
EDIT:
Thanks to @Somos in the comments. I was able to come up with my own solution. (Adding here if someone ever stumbles upon here looking for a solution.)
Subtract $\lfloor{x}\rfloor$ from both sides.
Then we need to prove that, $$-\lfloor{x}\rfloor+\bigg\lfloor{\frac{x+1}{2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2}{2^2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2^2}{2^3}}\bigg\rfloor+ \ldots=0$$
We know that, $\lfloor{x}\rfloor=\bigg\lfloor{\frac{x}{2}}\bigg\rfloor+\bigg\lfloor{\frac{x+1}{2}}\bigg\rfloor$
Therefore, $-\lfloor{x}\rfloor+\bigg\lfloor{\frac{x+1}{2}}\bigg\rfloor=-\bigg\lfloor{\frac{x}{2}}\bigg\rfloor$
Subsituting this,
$$-\bigg\lfloor{\frac{x}{2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2}{2^2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2^2}{2^3}}\bigg\rfloor+ \ldots$$
$$-\bigg\lfloor{\frac{x}{2}}\bigg\rfloor+\bigg\lfloor{\frac{\frac{x}{2}+1}{2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2^2}{2^3}}\bigg\rfloor+ \ldots$$
Similarly, I reach,
$$\lim_{n\to\infty}-\bigg\lfloor{\frac{x}{2^n}}\bigg\rfloor$$
i.e. $$=0$$
Hence Proved.