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