Tính không quy đồng mẫu:

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{32}-\frac{1}{64}\)

\(A=1-\frac{1}{64}=\frac{63}{64}\)

A = 32 + 16 + 8 + 4 + 2 + 1/64 = 63/64

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\(\dfrac{63}{64}\)

\(\dfrac{32}{64}+\dfrac{16}{64}+\dfrac{8}{64}+\dfrac{4}{64}+\dfrac{2}{64}+\dfrac{1}{64}=\dfrac{32+16+8+4+2+1}{64}=\dfrac{63}{64}\)

 1/2    + 1/4     + 1/8  + 1/16  + 1/32 + 1/64

 = 32/64 + 16/64 + 8/64 + 4/64 + 2/64 + 1/64

 =                         63/64

\(2A=1+\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\)

\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{18}+\dfrac{1}{32}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\right)\)

\(A=1-\dfrac{1}{64}\)

\(A=\dfrac{63}{64}\)

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= 63/64 nha

\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)

\(\dfrac{4}{2}A=\dfrac{4}{2}\cdot\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\right)\)

\(2A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\)

\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\right)\)

\(A=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+..\left(\dfrac{1}{32}-\dfrac{1}{32}\right)+\left(1-\dfrac{1}{64}\right)\)

\(A=1-\dfrac{1}{64}\)

\(A=\dfrac{63}{64}\)

\(\dfrac{127}{128}\)

A=21​+41​+91​+161​+321​+641​

42�=42⋅(12+14+18+116+132+164)24​A=24​⋅(21​+41​+81​+161​+321​+641​)

2�=1+12+14+18+116+1322A=1+21​+41​+81​+161​+321​

2�−�=(1+12+14+18+116+132)−(12+14+18+116+132+164)2A−A=(1+21​+41​+81​+161​+321​)−(21​+41​+81​+

Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)

\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)

\(A=1-\frac{1}{64}\)

\(A=\frac{63}{64}\)

=32/64+14/64+8/64+4/64+2/64+1/64

=32+14+8+4+2+1/64

=61/64

1/2 +1/4+ 1/8+1/16+1/32+1/64

=1/1×2+1/2×2+1/2×4+1/4×4+1/4+8+1/8×8

=1/1-1/2+1/2-1/2+1/2-1/4+1/4-1/4+1/4-1/8+1/8-1/8

=1-1/8

=7/8

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}\)

\(=1-\frac{1}{64}\)

\(=\frac{63}{64}\)

\(\frac{1}{2}\)+\(\frac{1}{4}\)+\(\frac{1}{8}\)+\(\frac{1}{16}\)+\(\frac{1}{32}\)+\(\frac{1}{64}\)=\(\frac{32}{64}\)+\(\frac{16}{64}\)+\(\frac{8}{64}\)+\(\frac{4}{64}\)+\(\frac{2}{64}\)+\(\frac{1}{64}\)

                                                                        =\(\frac{32+16+8+4+2+1}{64}\)

                                                                        =\(\frac{63}{64}\)

hok tốt

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}=\left(\frac{1}{2}+\frac{1}{8}\right)+\left(\frac{1}{4}+\frac{1}{16}\right)+\left(\frac{1}{32}+\frac{1}{64}\right)\)

\(=\left(\frac{4}{8}+\frac{1}{8}\right)+\left(\frac{4}{16}+\frac{1}{16}\right)+\left(\frac{2}{64}+\frac{1}{64}\right)=\frac{5}{8}+\frac{5}{16}+\frac{3}{64}\)

\(=\frac{40}{64}+\frac{20}{64}+\frac{3}{64}=\frac{40+20+3}{64}=\frac{63}{64}\)

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}\)

\(=1-\frac{1}{64}\)

\(=\frac{63}{64}\)

Ta có:\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\)\(\frac{1}{64}\)

     \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}\)\(+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\)\(\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}\)

     \(=1-\frac{1}{64}\)\(=\frac{63}{64}\)

Mình kết bạn nhé !

Cộng thêm 1/2 vào biểu thức đã cho, có:

S + 1/2= 1/2+1/4+ 1/8+ 1/16+1/32+1/64+1/128

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