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Question-186736




Question Number 186736 by Rupesh123 last updated on 09/Feb/23
Answered by Frix last updated on 09/Feb/23
⌊(2^0 /3)⌋=0 ⇒  A_(2n) =Σ_(i=1) ^(2n) ⌊(2^i /3)⌋=Σ_(i=1) ^n ⌊(2^(2i) /3)⌋+Σ_(i=1) ^n ⌊(2^(2i−1) /3)⌋  a_i =⌊(2^(2i) /3)⌋=⟨1, 5, 21, 65, ...⟩=((4^i −1)/3)  b_i =⌊(2^(2i−1) /3)⌋=⟨0, 2, 10, 42, ...⟩=((4^i −4)/6)  a_i +b_i =((4^i −2)/2)  A_(2n) =Σ_(i=1) ^n ((4^i −2)/2)=((2×4^n −3n−2)/3)  A_(100) =((2(4^(50) −76))/3)≈8.45×10^(29)
$$\lfloor\frac{\mathrm{2}^{\mathrm{0}} }{\mathrm{3}}\rfloor=\mathrm{0}\:\Rightarrow \\ $$$${A}_{\mathrm{2}{n}} =\underset{{i}=\mathrm{1}} {\overset{\mathrm{2}{n}} {\sum}}\lfloor\frac{\mathrm{2}^{{i}} }{\mathrm{3}}\rfloor=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\lfloor\frac{\mathrm{2}^{\mathrm{2}{i}} }{\mathrm{3}}\rfloor+\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\lfloor\frac{\mathrm{2}^{\mathrm{2}{i}−\mathrm{1}} }{\mathrm{3}}\rfloor \\ $$$${a}_{{i}} =\lfloor\frac{\mathrm{2}^{\mathrm{2}{i}} }{\mathrm{3}}\rfloor=\langle\mathrm{1},\:\mathrm{5},\:\mathrm{21},\:\mathrm{65},\:…\rangle=\frac{\mathrm{4}^{{i}} −\mathrm{1}}{\mathrm{3}} \\ $$$${b}_{{i}} =\lfloor\frac{\mathrm{2}^{\mathrm{2}{i}−\mathrm{1}} }{\mathrm{3}}\rfloor=\langle\mathrm{0},\:\mathrm{2},\:\mathrm{10},\:\mathrm{42},\:…\rangle=\frac{\mathrm{4}^{{i}} −\mathrm{4}}{\mathrm{6}} \\ $$$${a}_{{i}} +{b}_{{i}} =\frac{\mathrm{4}^{{i}} −\mathrm{2}}{\mathrm{2}} \\ $$$${A}_{\mathrm{2}{n}} =\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{4}^{{i}} −\mathrm{2}}{\mathrm{2}}=\frac{\mathrm{2}×\mathrm{4}^{{n}} −\mathrm{3}{n}−\mathrm{2}}{\mathrm{3}} \\ $$$${A}_{\mathrm{100}} =\frac{\mathrm{2}\left(\mathrm{4}^{\mathrm{50}} −\mathrm{76}\right)}{\mathrm{3}}\approx\mathrm{8}.\mathrm{45}×\mathrm{10}^{\mathrm{29}} \\ $$

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