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

Commented by bramlex last updated on 25/Jul/20

Σ_(i=0) ^∞ ⌊ ((n+2^i )/2^(i+1) )⌋   n=1 , ⌊((1+1)/2)⌋+⌊((1+2)/4)⌋+...= 1  n=2 , ⌊((2+1)/2)⌋+⌊((2+2)/4)⌋+...=2  n=3, ⌊((3+1)/2)⌋+⌊((3+2)/4)⌋+...=3  n=4 ,⌊((4+1)/2)⌋+⌊((4+2)/4)⌋+⌊((4+3)/8)⌋+...=4  Σ_(i=0) ^∞ ⌊((n+2^i )/2^(i+1) )⌋ = n , ∀n ∈N

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

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)

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

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

no sir  [...] mean integr parts(floor) you answer another question..!

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

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

Oh ! OK Thanks

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