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`1/2+1/4+1/8+...+1/128`
`=1/2x2+1/4x2+1/8x2+...+1/128x2`
`=1+1/2+1/4+1/8+...+1/64`
`=1+1/2+1/4+1/8+...+1/64-1/2-1/4-1/8-...-1/128`
`=1-1/128`
`=127/128`
Đặt A=\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(2A-A=1-\frac{1}{128}\)
\(A=\frac{127}{128}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=\frac{127}{128}\)
1/2+1/4+1/8+1/16+1/32+1/64+1/128
=64/128+32/128+16/128+8/126+4/126+2/126+1/128
=64+32+16+8+4+2+1/128
=123/128
C=1/2+1/4+1/8+1/16+1/32+1/64+1/128
C=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
C=1-1/128
C=127/128
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{128}\)
Đặt : \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{128}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{64}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\right)\)
\(\Rightarrow A=1-\frac{1}{128}\)
\(\Rightarrow A=\frac{127}{128}\)
Vậy : \(A=\frac{127}{128}\)
\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+...+\frac{1}{512}-\frac{1}{1024}\)
\(A=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^9}-\frac{1}{2^{10}}\)
\(2A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+...+\frac{1}{2^8}-\frac{1}{2^9}\)
\(3A=1-\frac{1}{2^{10}}< 1\)
\(\Rightarrow A< \frac{1}{3}\)
1+2+4+6+8+.........+=4160
Nếu làm đúng thì k cho mình nha thanh tam
gọi dãy số 1/2+1/4+1/8+...+1/128 là A
A=1/2+1/4+1/8+...+1/128
A=1/2+1/2^2+1/2^3+...+1/2^7
2A=1/2^2+1/^3+1/2^4+...+2^8
2A-A=A
ta có
1/2^2+1/2^3+1/2^4+...+1/2^8-(1/2+1/2^2+1/2^3+...+1/2^7)
=1/2^8-1/2
=Tự tìm ra nhé
=