\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{2048}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-...-\frac{1}{2048}+\frac{1}{2048}\)

\(=1-\frac{1}{2048}\)

\(=\frac{2047}{2048}\)

k mk nha

Đặt A = 1/2¹ + 1/2² + 1/2³ + 1/2⁴ +... +1/2¹¹

 2A = (1/2¹ + 1/2² + 1/2³ + 1/2⁴ +... +1/2¹¹⁾.2

2A =  1 + 1/2 +1/2² + ...+ 1/2¹⁰

2A - A = ( 1 + 1/2 +1/2² + ...+ 1/2¹⁰ ) -  (1/2¹ + 1/2² + 1/2³ + 1/2⁴ +... +1/2¹¹)

A = 1 - 1/2¹¹

A = 1 - 1/2048

A = 2047/2048

NHA CÁC BẠN ^_^

Đặt S = 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/2048 

   2.S = 1 + 1/2 + 1/4 + 1/8 + ... + 1/2048

2.S - S = ( 1 + 1/2 + 1/4 + 1/8 + ... + 1/2048 ) - ( 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/2048 )

      S  =  1 - 1/2048

       S =       2047/2048     

Trả lời bài này khó à nha

tất cả rút gọn = 1 /2

và GẠCH TỬ SỐ,MẪU SỐ GIỐNG NHAU 

kết quả = 1 nhé

Coi mk làm nha

1/2+1/4+1/8+....+1/2048

=1+1/2+1/4+1/8+.....+1/2048

=(1+1/2+.....+1/2048)-(1/2+1/4+...+1/2048)

=1-1/2048

=2047/2048

= 2047/2048 nha bạn k mình nha 

1024 phan 2048

A=1/2+1/4+1/8+1/16+...+1/2048

2A=1+1/2+1/4+1/1/8+...+1/1024

2A-A=(1+1/2+...+1/1024)-(1/2+1/4+...+1/2048)

A=1-1/2048

A=2047/2048

Đặt: \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}\)

\(\Rightarrow A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{11}}\)

\(\Rightarrow2A=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{11}}\right)\)

\(=1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{10}}\)

\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{11}}\right)\)

\(\Rightarrow A=1-\frac{1}{2^{11}}=\frac{2^{11}-1}{2^{11}}=\frac{2047}{2048}\)
 

\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\)

\(2A-A=\left(1+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+...+\frac{1}{2048}\right)\)

\(A=1-\frac{1}{2048}=\frac{2047}{2048}\)

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....+\frac{1}{2048}\)

\(=\left(1+\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2048}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}\right)\)

\(=1-\frac{1}{2048}\)

\(=\frac{2047}{2048}\)

gọi biểu thức là A

A=1/2+1/4+1/8+...+1/2048=1/2+1/2^2+1/2^3+...+1/2^10

=>2A=1+1/2+1/2^2+...+1/2^9

=>A=2A-A(bạn đặt cột dọc ra rồi sẽ thấy:1/2-1/2=0;1/2^2-1/2^2=0;...)Ta được kết quả bằng 1+1/2^10

Đặt A =1/2 + 1/4 + 1/8 + ...+ 1/1024 + 1/2048

A= 1/2 + 1/2^2 + 1/2^3+...+ 1/2^10 + 1/2^11

2A= 1 +1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10

2A-A= (1 +1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10) - (1/2 + 1/2^2 + 1/2^3+...+ 1/2^10 + 1/2^11)

A= 1+1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10 - 1/2 - 1/2^2 - 1/2^3 - ...- 1/2^10 - 1/2^11

A= 1- 1/2^11

A= 2047/ 2048

Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+...+\frac{1}{1024}+\frac{1}{2048}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+...+\frac{1}{512}+\frac{1}{1024}\)
\(2A-A=1-\frac{1}{2048}\)
\(A=\frac{2047}{2048}\)

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{2048}\)

\(=1-\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}\)

\(=1-\frac{1}{2048}\)

\(=\frac{2047}{2048}\)

 NHận xét :

1/2 = 1 - 1/2

1/4 = 1/2 - 1/4

1/8 = 1/4 - 1/8

...................

1/2048 = 1/1024 - 1/2048

Vậy 1/2 + 1/4 + 1/8 + ... + 1/2048

= 1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + ... - ... + 1/1024 - 1/2048

= 1 - 1/2048 = 2047/2048

Làm sao để ghi dấu phần vậy bạn

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+..+\frac{1}{2048}\)

\(=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+..+\frac{1}{2048}\)

\(=1-\frac{1}{2048}\)

\(=\frac{2047}{2048}\)

2047/2048 nha bạn

Ta có:
1/2 = 1 - 1/2
1/4 = 1/2 - 1/4
1/8 = 1/4 - 1/8
…
1/2048 = 1/1024 - 1/2048
Suy ra:
1/2 + 1/4 + 1/8 + ... + 1/2048
= 1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + ... - 1/1024 + 1/1024 - 1/2048
= 1 - 1/2048
= 2047/2048

k mình nha

Ta có :\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}+\frac{1}{2048}\)

nên \(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}+\frac{1}{512}+\frac{1}{1024}\)

Do đó :  \(2.A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}+\frac{1}{512}+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}+\frac{1}{2048}\right)\)

\(A=1-\frac{1}{2048}=\frac{2047}{2048}\)

Nhớ k giùm mình nhớ

ket qua là 2047\2048