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Ta có : \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{64}\)
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^6}\)
\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^5}\)
\(\Rightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^5}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^6}\right)\)
\(\Rightarrow A=1-\dfrac{1}{2^6}=1-\dfrac{1}{64}=\dfrac{63}{64}\)
\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+...+\dfrac{1}{32}-\dfrac{1}{64}\)
\(=1-\dfrac{1}{64}\)
\(=\dfrac{63}{64}\)
Đặt A = 1/2 − 1/4 + 1/8 − 1/16 + 1/32 − 1/64
A = 1/2 − 1/4 + 1/8 − 1/16 + 1/32 − 1/64
2A = 1 − 1/2 + 1/4 − 1/8 + 1/16 − 1/322
A =1 − 1/2 + 1/4 − 1/8 + 1/16 − 1/32
3A = 2A + A = 1 − 1/64 < 1
⇒ A < 1/3
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Tk cho mình nha
A=1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64
A= ( 1/2 - 1/4 ) + ( 1/8 - 1/16 ) + ( 1/32 - 1/64 )
A= 1/4 + 1/16 + 1/64
A = 16/64 + 4/64 + 1/64
A = 16+4+1/64
A= 21/64
Ta có : 1/3 = 21/63 mà 21/64 < 21/63 => 21/64 < 1/3 => 1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/ 64 < 1/3
Vậy 1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/ 64 < 1/3 ( đã chứng minh được )
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{32}-\frac{1}{64}\)
\(=\frac{1}{1}-\frac{1}{64}=\frac{63}{64}\)
a) \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{64}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{32}\)
\(\Rightarrow2A-A=A=1-\frac{1}{64}=\frac{63}{64}\)
=> \(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{32}\)
=> \(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{64}\right)\)
=> \(A=1+\frac{1}{32}-\frac{1}{16}-\frac{1}{64}=\frac{61}{64}\)
A = 1/2 + 1/4 + 1/8 + 1/16 + 1/64 = 32/64 + 16/64 + 8/64 + 4/64 + 1/64 = 61/64.