Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(2A=2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\right)\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+.......+\frac{1}{64}+\frac{1}{128}\right)-\left(\frac{1}{2}+\frac{1}{4}+......+\frac{1}{128}+\frac{1}{256}\right)\)
\(\Rightarrow A=1-\frac{1}{256}\)
\(\Rightarrow A=\frac{255}{256}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(=\frac{128}{256}+\frac{64}{256}+\frac{32}{256}+\frac{16}{256}+\frac{8}{256}+\frac{4}{256}+\frac{2}{256}+\frac{1}{256}\)
\(=\frac{\left(128+2\right)+\left(64+16\right)+\left(32+8\right)+\left(4+1\right)}{256}\)
\(=\frac{130+80+40+5}{256}\)
\(=\frac{255}{256}\)
\(A=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\)
\(\Rightarrow2A=1+\frac{1}{2}+...+\frac{1}{128}\)
\(\Rightarrow A=2A-A=1-\frac{1}{256}\)
\(\Rightarrow A=\frac{255}{256}\)
Đặt:\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)
\(\Rightarrow A=\frac{1}{2^1}+\frac{1}{2^2}+....+\frac{1}{2^8}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+..+\frac{1}{2^7}\right)-\left(\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^7}\right)\)
\(\Rightarrow A=1-\frac{1}{2^7}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)\(+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
= \(\left(\frac{1}{2}+\frac{1}{8}\right)+\left(\frac{1}{4}+\frac{1}{16}\right)+\left(\frac{1}{32}+\frac{1}{64}\right)\)\(+\left(\frac{1}{128}+\frac{1}{256}\right)\)
= \(\left(\frac{4}{8}+\frac{1}{8}\right)+\left(\frac{4}{16}+\frac{1}{16}\right)+\left(\frac{2}{64}+\frac{1}{64}\right)\) \(+\left(\frac{2}{256}+\frac{1}{256}\right)\)
= \(\frac{5}{8}+\frac{5}{16}+\frac{3}{64}+\frac{3}{256}\)
= \(\frac{160}{256}+\frac{80}{256}+\frac{12}{256}+\frac{3}{256}\)
= \(\frac{255}{256}\)
^^ Ủng hộ nha!
1 - 1/256 = \(\frac{255}{256}\)
Kết quả đấy mình thử lại đúng luôn
=(1-\(\frac{1}{2}\))+(\(\frac{1}{2}\)-\(\frac{1}{4}\))+(\(\frac{1}{4}\)-\(\frac{1}{8}\))+....+(\(\frac{1}{128}\)\(\frac{1}{256}\))
=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{8}\)+....+\(\frac{1}{128}\)-\(\frac{1}{256}\)
=1-\(\frac{1}{256}\)=\(\frac{255}{256}\)
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}+\frac{1}{2^8}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}\)
\(\Rightarrow2A-A=1-\frac{1}{2^8}=\frac{255}{256}\)
\(\Rightarrow A=\frac{255}{256}\)
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(2A=\frac{1}{2}\times2+\frac{1}{4}\times2+\frac{1}{8}\times2+\frac{1}{16}\times2+...+\frac{1}{256}\times2\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\right)\)
\(A=1-\frac{1}{256}\)
\(A=\frac{255}{256}\)
\(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)
\(2B=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\)
\(2B-B=\left(1+\frac{1}{2}+...+\frac{1}{128}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)\)
\(2B-B=1-\frac{1}{256}\)
\(B=\frac{255}{256}\)
=1-1/2+1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32+1/32-1/64+1/64-1/128+1/128-1/256
=1-1/256
=1
Trl
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Hok tốt
~ NHÉ BẠN ~
