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

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

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

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

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

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

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

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

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​+161​+321​+641​)

�=(12−12)+(14−14)+..(132−132)+(1−164)A=(21​−21​)+(41​−41​)+..(321​−321​)+(1−641​)

�=1−164A=1−641​

�=6364A=6463​
 

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

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

 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

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

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

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

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

giúp em với ạ , em cấn gấp ,  4h em học rồi ạ . Cảm ơn những ai giúp em ạ , em tick luôn ạ

= 63/64 nha

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

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

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

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

\(\Rightarrow2A-A=A=1-\frac{1}{64}=\frac{63}{64}\)

b: A=1/3+1/9+...+1/3^10

=>3A=1+1/3+...+1/3^9

=>A*2=1-1/3^10=(3^10-1)/3^10

=>A=(3^10-1)/(2*3^10)

c: C=3/2+3/8+3/32+3/128+3/512

=>4C=6+3/2+...+3/128

=>3C=6-3/512

=>C=1023/512

d: A=1/2+...+1/256

=>2A=1+1/2+...+1/128

=>A=1-1/256=255/256

😵

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

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

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

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

        A       =  1 - \(\dfrac{1}{32}\)

        A       =   \(\dfrac{31}{32}\)

      B = \(\dfrac{1}{3}\) + \(\dfrac{1}{9}\) + \(\dfrac{1}{27}\) + \(\dfrac{1}{81}\) + \(\dfrac{1}{243}\) + \(\dfrac{1}{729}\)

3 \(\times\)B =  1 + \(\dfrac{1}{3}\) +   \(\dfrac{1}{9}\) +  \(\dfrac{1}{27}\) + \(\dfrac{1}{81}\) + \(\dfrac{1}{243}\)

3 \(\times\) B - B = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{9}\) + \(\dfrac{1}{27}\) + \(\dfrac{1}{81}\) + \(\dfrac{1}{243}\) -(\(\dfrac{1}{3}\)+\(\dfrac{1}{9}\)+\(\dfrac{1}{27}\)+...+\(\dfrac{1}{729}\))

2B          = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{9}\) + \(\dfrac{1}{27}\) + \(\dfrac{1}{81}\) + \(\dfrac{1}{243}\) - \(\dfrac{1}{3}\) - \(\dfrac{1}{9}\)-\(\dfrac{1}{27}\)-...- \(\dfrac{1}{729}\)

2B          = 1 - \(\dfrac{1}{729}\)

2B         = \(\dfrac{728}{729}\)

   B = \(\dfrac{728}{729}\) : 2

   B  = \(\dfrac{364}{729}\)

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

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

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