s=1/2¹ + 1/2²+1/2³+.........+1/2¹⁰

2s=  1/2²   +1/2³+.........+1/2¹⁰+1/2¹¹

2s-s= ( 1/2²   +1/2³+.........+1/2¹⁰+1/2¹¹ ) -(1/2¹ + 1/2²+1/2³+.........+1/2¹⁰)

s= 1/2¹¹        -1/2

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}\)

\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}\right)\)

\(A=1-\frac{1}{512}=\frac{511}{512}\)

Đặt:  \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)

\(\Rightarrow2A=2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\right)\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)

\(\Rightarrow2A-A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)\(-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\right)\)

\(\Rightarrow A=1-\frac{1}{512}\)

\(\Rightarrow A=\frac{511}{512}\)

~ rất vui vì giúp đc bn ~

ta gọi tổng trên là H

  Ta có:   H=1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256

⇒2H=1+1/2+1/4+1/8+1/16+1/32+1/64+1/128

⇒2H−H=(1+1/2+1/4+.........+1/128)−(1/2+1/4+1/8+...........+1/256)

=>H= 1 + 1/2+1/4+1/8+1/16+1/32+1/64-1/2-1/4-/18-1/16-1/32-1/64-1/128-1/256

⇒H=1−1/256

⇒H=255/256

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

Ta thấy:

1/2=1/1-1/2

1/4=1/2-1/4

1/8=1/4-1/8....

1/64=1/32-1/64

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

A=1 - 1/ 64

A= 63/64

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

ta thấy:

1/2=1/1-1/2

1/4=1/2-1/4

1/8=1/4-1/8...

1/64=1/32-1/64

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

A=1/1-1/64

A=63/64

a = 1/(1×2) + 1/(2×3) + 1/(3×4) + 1/(5×6)

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6

= 1 - 1/6

= 5/6

cảm ơn bn nhé

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

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

Đặt A=1/2+1/4+...+1/128

=1/2+(1/2)^2+...+(1/2)^7

=>2A=1+1/2+...+(1/2)^6

=>2A-A=1+1/2+...+(1/2)^6-1/2-1/4-...-1/128

=>A=1-1/128=127/128