Ta có:

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

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

\(\Rightarrow2A-A=A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)

\(=1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{10}}=\frac{1023}{1024}\)

(^oy^o)

a=1/2+1/2^2+.........+1/2^10

2a=1+1/2+1/2^2+........+1/2^9

2a-a=1-2^10

=1023/1024

2A=2*(1/2+1/2^2+....+1/2^10)

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

2A - A =A=1 - 1/2^10

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

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

Lấy (2) - (1) ta có ;

=> A = 1-1/2^10

Vậy.................

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

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

2A - A = 1 - 1/2^10 

   A     = 1 - 1/2^10 

Chúc học giỏi !!! 

\(A=\dfrac{\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}}{\dfrac{2}{2}+\dfrac{2}{2^2}+...+\dfrac{2}{2^{10}}}=\dfrac{\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}}{2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\right)}=\dfrac{1}{2}\)

A = \(\dfrac{\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{10}}}{\dfrac{2}{2}+\dfrac{2}{2^2}+\dfrac{2}{2^3}+...+\dfrac{2}{2^{10}}}\)

= \(\dfrac{\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}}{2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)}\)

= \(\dfrac{1}{2}\)

S/2=1/2²+1/2³+1/3⁴+...+1/2¹¹

suy ra S-S/2=S/2=1/2-1/2¹¹

suy ra S=1-1/2¹⁰=1023/1024

(1+10) +(2+9) +....+ (5+6) + (10+1) +(9+2) +........+ (6+5)

= 11 x 10 = 110

Giải

 (1+9)+(2+8)...(6+4)+(7+3)...(9+1)+10+10

10*11=110