Từ biểu thức trên ta được

A=1/2+1/2^2+1/2^3+....+1/2^11+1/2^12

2A-A=2(1/2+1/2^2+1/2^3+....+1/2^11+1/2^12)-(1/2+1/2^2+1/2^3+....+1/2^11+1/2^12)

A=1+1/2^2+1/2^3+....+1/2^11-1/2-1/2^2-...-1/2^12

A=1/2-1/2^12

Ủng hộ cho mình nha bạn

\(A=\)\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}+\frac{1}{2048}+\frac{1}{4096}\)

\(A=\)\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{12}}\)

\(2A=\)\(2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{12}}\right)\)

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

\(2A-A=\)\(\left(1+\frac{1}{2}+...+\frac{1}{2^{11}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{12}}\right)\)

\(A=1-\frac{1}{2^{12}}\)

Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}+\frac{1}{2048}+\frac{1}{4096}\)

\(\Leftrightarrow A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{10}}+\frac{1}{2^{11}}+\frac{1}{2^{12}}\)

Nhân 2 vào 2 vế của biểu thức A , ta được :

\(2A=2\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{10}}+\frac{1}{2^{11}}+\frac{1}{2^{12}}\right)\)

\(\Rightarrow2A=1+\frac{1}{2^1}+\frac{1}{2^2}+....+\frac{1}{2^9}+\frac{1}{2^{10}}+\frac{1}{2^{11}}\)

Lấy biểu thức 2A - A , Ta được :

\(2A-A=\left(1+\frac{1}{2^1}+\frac{1}{2^2}+....+\frac{1}{2^{10}}+\frac{1}{2^{11}}\right)-\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{11}}+\frac{1}{2^{12}}\right)\)

\(\Rightarrow A=1-\frac{1}{2^{12}}\)

2

4

8

16

32

64

128

256

512

1024

2048

4096

8192

1+1=2

2+2=4

4+4=8

8+8=16

16+16=32

64+64=128

128+128=256

512+512=1024

2048+2048=4096

xong

1+1=2

2+2=4

4+4=8

8+8=16

16+16=32

32+32=64

64+64=128

128+128=256

256+256=512

512+512=1024

1024+1024=2048

2048+2048=4096

4096+4096=8192

Đặt \(A=1+2+4+.........+4096\)

\(2A=2+4+8+......+8192\)

\(\Rightarrow2A-A=8192-1\)

\(\Rightarrow A=8191\)

Đặt  \(S=1+2+4+...+1024+2048+4096\)

       \(S=1+2^1+2^2+2^3+....+2^{10}+2^{11}+2^{12}\)

    \(2S=2+2^2+2^3+....+2^{11}+2^{12}+2^{13}\)

\(2S-S=\left(2+2^2+2^3+....+2^{12}+2^{13}\right)-\left(1+2+2^2+....+2^{11}+2^{12}\right)\)

     \(S=2^{13}-1=8192-1=8191\)

            A=1+2+4+8+16+.....+1024+2048+4096

          2A= 2+4+8+16+32+....+2048+4096+8192

  2A-A=A=2+4+8+16+32+....+4096+8192-(1+2+4+8+16+...+1024+2048+4096)

           A=8192-1

            A=8191 k nha 

Đặt 

\(S=1+2+4+...+2048+4096\)

\(S=1+2^1+2^2+...+2^{11}+2^{12}\)

\(2S=2+2^2+2^3+...+2^{12}+2^{13}\)

\(2S-S=\left(2+2^2+2^3+...+2^{13}\right)-\left(1+2+2^2+..+2^{12}\right)\)

\(S=2^{13}-1=8192-1=8191\)

Gọi A=1+2+4+8+16+...+1024+2048+4096

2A=2+4+8+16+32+...+2048+4096+8192

2A-A=(2+4+8+16+32+...+2048+4096+8192)-(1+2+4+8+16+...+1024+2048+4096)

A=8192-1

A=8191

                Đặt A = 1 + 2 + 4 + 8 + 16 +...+ 1024 + 2048 + 4096

                A = 1 + 2 + 2² + 2³ + 2⁴ +.... + 2¹⁰ + 2¹¹ + 2¹²

                     2A = 2 + 2² + 2³ + 2⁴ + 2⁵ + ..... + 2¹¹ + 2¹² + 2¹³

               2A - A = (2 + 2² + 2³ + 2⁴ + .... + 2¹¹ + 2¹² + 2¹³) - (1 + 2 + 2² + 2³ + 2⁴ +.....+ 2¹⁰ + 2¹¹ + 2¹²)

              => A = 2¹³ - 1

              => A = 8192 - 1 = 8191

           Ủng hộ mk nha !!! ^_^

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https://olm.vn/hoi-dap/question/656309.html

https://olm.vn/hoi-dap/question/656309.html

Đặt A=1/2+1/4+1/8+1/16+1/32+...+1/2048+1/4096

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{12}}\)

\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{12}}\right)\)

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

\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^{11}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{12}}\right)\)

\(A=1-\frac{1}{2^{12}}\)

Ai muốn kết bn ko!

Tiện thể tk mình luôn nha!

Konasuba

Ta co: 2S=2+1+1/2+1/4+...+1/2048

          2S-S=2+1+1/2+1/4+...+1/2048-1-1/2-1/4-...-1/2048-1/4096

         \(\Rightarrow\)S=2-1/4096 =8191/4096

cảm ơn bạn nha

4095/4096