Đặt tổng trên = A

\(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{4096}+\frac{1}{8192}\)

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

\(A.2-A=\left(2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2048}+\frac{1}{4096}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{4096}+\frac{1}{8192}\right)\)

\(A=2-\frac{1}{8192}=\frac{16383}{8192}\)
 

Đặt A = 1 + 1/2 + 1/4 + 1/8 + ... + 1/4096 + 1/8192

2A = 2 + 1 + 1/2 + 1/4 + ... + 1/2048 + 1/4096

2A - A = (2 + 1 + 1/2 + 1/4 + ... + 1/2048 + 1/4096) - (1 + 1/2 + 1/4 + 1/8 +... + 1/4096 + 1/8192)

A = 2 - 1/8192

A = 16383/8192

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

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

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

A=\(1-\frac{1}{4096}\)

A=\(\frac{4095}{4096}\)

4095/4096 đó lam lợn hả

bn tham khảo tại link này nhé :

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

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

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=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

Bé ơi nghĩ chị bài toàn của bé giải như sau 
Tính nhanh tổng sau A = 1+2+4+8+16+32 +.....+ 8192 
****************************************... 
Đặt A × 2 = 2 + 4 +8 +16 + 32 + ....+ 16384 
Cùng thêm 1 và bớt 1 ta có như sau: 
A × 2 = 1 + 2 + 4 + 8 + 16 + .....+ 1892 + 16384 -1 
A × 2 = A + 16384 - 1 
A = 16384 -1 
A = 16383 
****************************************... 
Chúc bé học giỏi nha

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