1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256 = ?
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\(E=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{256}\)
\(2\times E=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}\)
\(2\times E-E=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{256}\right)\)
\(E=1-\dfrac{1}{256}\)
\(E=\dfrac{256}{256}-\dfrac{1}{256}\)
\(E=\dfrac{255}{256}\)
Tính \(S=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\)
Dùng sai phân như sau
\(2S-S=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)=1-\frac{1}{256}\)
Vậy \(S=1-\frac{1}{256}\)
\(\text{Đặt }\)\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(\Rightarrow2A-A=1-\frac{1}{256}\)
\(=>A=\frac{255}{256}\)
1 / 2 + 1 / 4 + 1 / 8 + 1 / 16 + 1 / 32 + 1 / 64 + 1 / 128 = 127 / 128
A= 1/2 + 1/4+ 1/8+ 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512
A = 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 - 1/256 - 1/512
A = 1 - 1/512
A = 511/512
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gọi biểu thức đó là A
A=1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256+1/512
1/512+A=1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256+1/512+1/512
1/512+A=1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256+1/256
1/512+A=1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/128
1/512+A=1/2+1/4+1/8+1/16+1/32+1/64+1/64
1/512+A=1/2+1/4+1/8+1/16+1/32+1/32
1/512+A=1/2+1/4+1/8+1/16+1/16
1/512+A=1/2+1/4+1/8+1/8
1/512+A=1/2+1/4+1/4
1/512+A=1/2+1/2
1/512+A=1
A=1-1/512
A=511/512
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