\(y=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}=?\)
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\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}\)
\(=1-\frac{1}{32}\)
\(=\frac{31}{32}\)
có tử = 1 thì k bt à nha
Đặt: \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(\Rightarrow\)\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
\(\Rightarrow\)\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\)
\(\Rightarrow\)\(A=1-\frac{1}{32}=\frac{31}{32}\)
mà \(A=\frac{1}{x}\)
nên \(\frac{1}{x}=\frac{31}{32}\)
\(\Rightarrow\)\(x=\frac{32}{31}\)
Vậy...
= 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
= 1-1/128
=127/128
\(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{8}\)+ \(\frac{1}{16}\)+ \(\frac{1}{32}\)+ \(\frac{1}{64}\)+ \(\frac{1}{128}\)= \(\frac{64}{128}\)+ \(\frac{32}{128}\)+ \(\frac{16}{128}\)+ \(\frac{8}{128}\)+ \(\frac{4}{128}\)+ \(\frac{2}{128}\)+ \(\frac{1}{128}\).
= \(\frac{127}{128}\).
\(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{8}\)+ \(\frac{1}{16}\)+ \(\frac{1}{32}\)+ \(\frac{1}{64}\)+ \(\frac{1}{128}\)
= \(1\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{8}\)+ \(\frac{1}{8}\)- \(\frac{1}{16}\)+ \(\frac{1}{16}\)- \(\frac{1}{32}\)+ \(\frac{1}{32}\)- \(\frac{1}{64}\)+ \(\frac{1}{64}\)- \(\frac{1}{128}\)
= \(1\)- \(\frac{1}{128}\)
= \(\frac{127}{128}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}+\frac{1}{64}-\frac{1}{128}\)
\(=1-\frac{1}{128}\)
\(\frac{127}{128}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{128}-\frac{1}{256}\)
=\(1-\frac{1}{256}\)
=\(\frac{255}{256}\)
1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256
= 128/256 + 64/256 + 32/256 + 16/256 + 8/256 + 4/256 + 2/128 + 1/256
= 255/256
\(y=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(\frac{1}{2}y=\frac{1}{2}\times\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\)
\(\frac{1}{2}y=\frac{1}{2}\times1+\frac{1}{2}\times\frac{1}{2}+...+\frac{1}{2}\times\frac{1}{32}\)
\(\frac{1}{2}y=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(y-\frac{1}{2}y=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+..+\frac{1}{64}\right)\)
\(\left(1-\frac{1}{2}\right)y=1-\frac{1}{64}\)
\(\frac{1}{2}y=\frac{63}{64}\)
\(y=\frac{63}{64}\div\frac{1}{2}=\frac{63}{32}\)