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Đặt A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256
=> 2A = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
=> 2A - A = (1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128) - (1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256)
=> A = 1 - 1/256
=> A = 255/256
Vậy: ...
Đặt tổng là A
\(2xA=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{64}+\dfrac{1}{128}\)
\(\Rightarrow A=2xA-A=1-\dfrac{1}{256}=\dfrac{255}{256}\)
Đặt \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}+\frac{1}{256}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{128}-\frac{1}{256}\)
\(\Rightarrow A=1-\frac{1}{256}\)
\(\Rightarrow A=\frac{255}{256}\)
Ta có : \(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}+\frac{1}{2^8}\)
\(\Rightarrow2A=1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^6}+\frac{1}{2^7}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^6}+\frac{1}{2^7}\right)-\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}+\frac{1}{2^8}\right)\)
\(\Rightarrow A=1-\frac{2}{8}=\frac{256}{256}-\frac{1}{256}=\frac{255}{256}\)
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 \(S=1-\frac{1}{256}\)
a) 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
b) B = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
3B = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243
3B - B = 1 - 1/729
2B = 728/729
B = 364/729
a) 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
b) B = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
3B = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243
3B - B = 1 - 1/729
2B = 728/729
B = 364/729
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}+\frac{1}{256}\)
\(Ax2=2x\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}+\frac{1}{256}\right)\)
\(Ax2=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\)
\(Ax2-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\right)-\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{128}-\frac{1}{256}\right)\)
\(A=1-\frac{1}{256}\)
\(A=\frac{255}{256}\)