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Question Number 104940 by 175mohamed last updated on 24/Jul/20

Commented by bramlex last updated on 25/Jul/20

$$\underset{i=0}{\sum ^{\mathrm{\infty }}}\lfloor \frac{n+2^i}{2^{i+1}}\rfloor$$$$n=1,\lfloor \frac{1+1}{2}\rfloor +\lfloor \frac{1+2}{4}\rfloor +...=1$$$$n=2,\lfloor \frac{2+1}{2}\rfloor +\lfloor \frac{2+2}{4}\rfloor +...=2$$$$n=3,\lfloor \frac{3+1}{2}\rfloor +\lfloor \frac{3+2}{4}\rfloor +...=3$$$$n=4,\lfloor \frac{4+1}{2}\rfloor +\lfloor \frac{4+2}{4}\rfloor +\lfloor \frac{4+3}{8}\rfloor +...=4$$$$\underset{i=0}{\sum ^{\mathrm{\infty }}}\lfloor \frac{n+2^i}{2^{i+1}}\rfloor =n,\mathrm{\forall }n\in \mathbb{N}$$$$$$

Answered by Ar Brandon last updated on 25/Jul/20

$${\mathrm{A}}_i=\underset{i=0}{\sum ^{\mathrm{\infty }}}\left[\frac{\mathrm{n}+2^i}{2^{i+1}}\right]=\underset{i=0}{\sum ^{\mathrm{\infty }}}\{\frac{\mathrm{n}}{2^{i+1}}+\frac{1}{2}\}$$$$=\frac{\mathrm{n}}{2}\underset{i=0}{\sum ^{\mathrm{\infty }}}\frac{1}{2^i}+\underset{i=0}{\sum ^{\mathrm{\infty }}}\frac{1}{2}$$

Commented by mathmax by abdo last updated on 25/Jul/20

$$\mathrm{no}\mathrm{sir}[...]\mathrm{mean}\mathrm{integr}\mathrm{parts}\left(\mathrm{floor}\right)\mathrm{you}\mathrm{answer}\mathrm{another}\mathrm{question}..!$$

Commented by Ar Brandon last updated on 25/Jul/20

Oh ! OK Thanks

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