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Question Number 186736 by Rupesh123 last updated on 09/Feb/23

Answered by Frix last updated on 09/Feb/23

$$\lfloor \frac{2^0}{3}\rfloor =0\Rightarrow$$$$A_{2n}=\underset{i=1}{\sum ^{2n}}\lfloor \frac{2^i}{3}\rfloor =\underset{i=1}{\sum ^n}\lfloor \frac{2^{2i}}{3}\rfloor +\underset{i=1}{\sum ^n}\lfloor \frac{2^{2i-1}}{3}\rfloor$$$$a_i=\lfloor \frac{2^{2i}}{3}\rfloor =⟨1,5,21,65,...⟩=\frac{4^i-1}{3}$$$$b_i=\lfloor \frac{2^{2i-1}}{3}\rfloor =⟨0,2,10,42,...⟩=\frac{4^i-4}{6}$$$$a_i+b_i=\frac{4^i-2}{2}$$$$A_{2n}=\underset{i=1}{\sum ^n}\frac{4^i-2}{2}=\frac{2\times 4^n-3n-2}{3}$$$$A_{100}=\frac{2(4^{50}-76)}{3}\approx 8.45\times {10}^{29}$$

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