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

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

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

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

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

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

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

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

đặt `A= 1/2 + 1/4 + 1/8 + 1/16+ 1/32 + 1/64`

`=> 2A = 2(1/2 + 1/4 + 1/8 + 1/16+ 1/32 + 1/64)`

`2A = 1+1/2 +1/4 +1/8+1/16 +1/32`

`=>A =2A -A =1+1/2 +1/4 +1/8+1/16 +1/32-1/2-1/4-1/8-1/16-1/32 -1/64`

`A = 1-1/64 = 64/64 -1/64 =63/64`

Lời giải:

$A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}$

$2\times A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}$

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

$A=1-\frac{1}{64}=\frac{63}{64}$

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

 2A = 2.(1/2+1/4+1/8+1/16+1/32+1/64)
            = 1 + 1/2+1/4+1/8+1/16+1/32
=> 2A - A = (1+1/2+1/4+1/8+1/16+1/32) - (1/2+1/4+1/8+1/16+1/32+1/64)
=> A = 1 - 1/64
         = 63/64

Mình kết bạn nhé !

Dấu ... có nghĩ là còn nữa, bài toán này ko có phân số cuối cùng

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)

\(=1-\frac{1}{2}+...+\frac{1}{128}=1-\frac{1}{128}=\frac{127}{128}\)

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

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

=(4/8+2/8+1/8)+(4/64+2/64+1/64)

=7/8+7/64

=56/64+7/64

=63/64

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

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

2 x B - B = 1 - \(\dfrac{1}{64}\)

           B = \(\dfrac{63}{64}\)

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}+\frac{1}{64}-\frac{1}{128}\)

\(=1-\frac{1}{128}\)

\(=\frac{127}{128}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+.........+\frac{1}{64}-\frac{1}{128}=\frac{1}{1}-\frac{1}{128}=\frac{127}{128}\)

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

A = 1 - 1/64 

A = 63/64

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

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

A = 1 - 1/64

A = 63/64