Question Number 104940 by 175mohamed last updated on 24/Jul/20
Commented by bramlex last updated on 25/Jul/20
$$\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\lfloor\:\frac{{n}+\mathrm{2}^{{i}} }{\mathrm{2}^{{i}+\mathrm{1}} }\rfloor\: \\ $$$${n}=\mathrm{1}\:,\:\lfloor\frac{\mathrm{1}+\mathrm{1}}{\mathrm{2}}\rfloor+\lfloor\frac{\mathrm{1}+\mathrm{2}}{\mathrm{4}}\rfloor+…=\:\mathrm{1} \\ $$$${n}=\mathrm{2}\:,\:\lfloor\frac{\mathrm{2}+\mathrm{1}}{\mathrm{2}}\rfloor+\lfloor\frac{\mathrm{2}+\mathrm{2}}{\mathrm{4}}\rfloor+…=\mathrm{2} \\ $$$${n}=\mathrm{3},\:\lfloor\frac{\mathrm{3}+\mathrm{1}}{\mathrm{2}}\rfloor+\lfloor\frac{\mathrm{3}+\mathrm{2}}{\mathrm{4}}\rfloor+…=\mathrm{3} \\ $$$${n}=\mathrm{4}\:,\lfloor\frac{\mathrm{4}+\mathrm{1}}{\mathrm{2}}\rfloor+\lfloor\frac{\mathrm{4}+\mathrm{2}}{\mathrm{4}}\rfloor+\lfloor\frac{\mathrm{4}+\mathrm{3}}{\mathrm{8}}\rfloor+…=\mathrm{4} \\ $$$$\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\lfloor\frac{{n}+\mathrm{2}^{{i}} }{\mathrm{2}^{{i}+\mathrm{1}} }\rfloor\:=\:{n}\:,\:\forall{n}\:\in\mathbb{N} \\ $$$$ \\ $$
Answered by Ar Brandon last updated on 25/Jul/20
![A_i =Σ_(i=0) ^∞ [((n+2^i )/2^(i+1) )]=Σ_(i=0) ^∞ {(n/2^(i+1) )+(1/2)} =(n/2)Σ_(i=0) ^∞ (1/2^i )+Σ_(i=0) ^∞ (1/2)](https://www.tinkutara.com/question/Q104965.png)
$$\mathrm{A}_{{i}} =\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\frac{\mathrm{n}+\mathrm{2}^{{i}} }{\mathrm{2}^{{i}+\mathrm{1}} }\right]=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left\{\frac{\mathrm{n}}{\mathrm{2}^{{i}+\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{2}}\right\} \\ $$$$\:\:\:\:\:=\frac{\mathrm{n}}{\mathrm{2}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{i}} }+\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by mathmax by abdo last updated on 25/Jul/20
![no sir [...] mean integr parts(floor) you answer another question..!](https://www.tinkutara.com/question/Q105020.png)
$$\mathrm{no}\:\mathrm{sir}\:\:\left[…\right]\:\mathrm{mean}\:\mathrm{integr}\:\mathrm{parts}\left(\mathrm{floor}\right)\:\mathrm{you}\:\mathrm{answer}\:\mathrm{another}\:\mathrm{question}..! \\ $$
Commented by Ar Brandon last updated on 25/Jul/20
Oh ! OK Thanks