\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}+\dfrac{1}{1024}\)

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

\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)

\(\Rightarrow2A-A=A=1-\dfrac{1}{2^{10}}\)

đây ko phải lớp 5 đúng ko ?

Đặt A=1/2+1/4+1/8+..+1/1024

Ax2=1+1/2+1/4+1/8+..+1/512( Nhân cả 2 vế với 2)

Ax2-A=(1+1/2+1/4+1/8+..+1/512)-(1/2+1/4+1/8+..+1/1024)

<=>A=1-1/1024

<=>A=1023/1024

Vậy biểu thức đã cho = 1023/1024

Đặt \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{256}+\dfrac{1}{512}\)

\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}+\dfrac{1}{256}\)

\(\Rightarrow A=2A-A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}+\dfrac{1}{256}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-...-\dfrac{1}{256}-\dfrac{1}{512}\)

\(\Rightarrow A=1-\dfrac{1}{512}=\dfrac{511}{512}\)

Đặt ⇒2A=1+12+14+...+1128+1256⇒2A=1+12+14+...+1128+1256

⇒A=1−1512=511512

\(=\left(2+4+6+...+98\right)\left(6-6\right)\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{512}\right)\)

=0

= ( 2+4+6+...+98 ) ( 6- 6) ( 1/2+1/4 + .......+ 1/ 512 ) 

= 0

Chúc bạn học tốt 

= (2 + 4 + 6 + ... + 98) x ( 6 - 6 ) x ( \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) +...+ \(\dfrac{1}{512}\))

= (2 + 4 + 6 + ... + 98) x      0      x ( \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) +...+ \(\dfrac{1}{512}\))

=  0

*****Vì số nào nhân với 0 cũng bằng 0

Ta có:

\(\dfrac{1}{2^2}=\dfrac{1}{2\cdot2}< \dfrac{1}{1\cdot2}\)

\(\dfrac{1}{3^2}=\dfrac{1}{3\cdot3}< \dfrac{1}{2\cdot3}\)

\(\dfrac{1}{4^2}=\dfrac{1}{4\cdot4}< \dfrac{1}{3\cdot4}\)

...

\(\dfrac{1}{9^2}=\dfrac{1}{9\cdot9}< \dfrac{1}{8\cdot9}\)

\(\dfrac{1}{10^2}=\dfrac{1}{10\cdot10}< \dfrac{1}{9\cdot10}\)

\(\Rightarrow A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{10^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)

\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(\Rightarrow A< 1-\dfrac{1}{10}\)

\(\Rightarrow A< \dfrac{9}{10}\)

\(\Rightarrow A< 1\) (vì: \(\dfrac{9}{10}< 1\))

221​=2⋅21​<1⋅21​

132=13⋅3<12⋅3321​=3⋅31​<2⋅31​

142=14⋅4<13⋅4421​=4⋅41​<3⋅41​

...

192=19⋅9<18⋅9921​=9⋅91​<8⋅91​

1102=110⋅10<19⋅101021​=10⋅101​<9⋅101​

⇒�=122+132+142+...+1102<11⋅2+12⋅3+13⋅4+...+19⋅10⇒A=221​+321​+421​+...+1021​<1⋅21​+2⋅31​+3⋅41​+...+9⋅101​

⇒�<1−12+12−13+...+19−110⇒A<1−21​+21​−31​+...+91​−101​

⇒�<1−110⇒A<1−101​

⇒�<910⇒A<109​

$\Rightarrow A<1$ (vì: 910<1109​<1)

ta có   A = 1/21 + 1/22 + 1/23 + 1/24 + ... + 1/40  > 1/40 + 1/40 +....+ 1/40 ( có 20 số hạng 1/40)
              = 20/40
              =1/2
      =) A> 1/2   (1)
  ta lại có  A = 1/21 + 1/22 + 1/23 + 1/24 + ... + 1/40 < 1/20 + 1/20 +...+ 1/20 ( có 20 số hạng 1/20)
                    =20/20
                    =1
       =) A <1 (2)
từ (1), (2) = 1/2 <A<1

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