\(\Leftrightarrow2-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{11}}\right)>0\)

Ta có: \(\frac{1}{2^{12}}-1=\left(\frac{1}{2}-1\right)\left(\frac{1}{2^{11}}+\frac{1}{2^{10}}+\frac{1}{2^9}+...+\frac{1}{2}+1\right)\)

\(\Rightarrow1+\frac{1}{2}+...+\frac{1}{2^{11}}=2\left(1-\frac{1}{2^{12}}\right)=2-\frac{1}{2^{11}}\)

\(\Rightarrow2-\left(1+\frac{1}{2}+...+\frac{1}{2^{11}}\right)=2-\left(2-\frac{1}{2^{11}}\right)=\frac{1}{2^{11}}>0\left(đpcm\right)\)

1-1/2-1/2^2-......-1/2^11

ta có:1-1/2-1/2^2-.....-1/2^11=1-(1/2+1/2^2+....+1/2^11)

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

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

2A=2.1/2+2.1/2^2+....+2.1/2^11

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

A=1-1/2^11=2048/2048-1/2048=2047/2048

vì 1-(1/2+1/2^2+1/2^3+...+1/2^11)=1-A

=> 1-(1/2+1/2^2+1/2^3+...+1/2^11)=1-2047/2048=2048/2048-2047/2048=1/2048=1/2^11

vậy 1-1/2-1/2^2-1/2^3-...-1/2^11=1/2^11

Ta có \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{10^2}< \dfrac{1}{9.10}\)

cộng vế với vê sta đc 

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

Vậy ta có đpcm 

A la dat tren tong

We have: A = 1/2 ^ 2 + 1/3 ^ 2 + 1/4 ^ 2 + ........... + 1/10 ^ 2

A = 1 / 2.2 + 1 / 3.3 + 1 / 4.4 + ....... + 1 / 10:10

A  <1 / 1.2 + /2.3 + 1/3.4 +......+1/9.10

A < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ....+ 1/9 - 1/10

A < 1-1/10

Ma 1 - 1/10 = 9/10 < 1

=>A < 1 (dpcm)

dễ 

1/2^2=1/1.2

1/3^2=1/2.3

1/4^2=1/3.4

....

1/10^2=1/9.10

1/1.2+1/2.3+1/3.4+...+1/9.10

=(1-1/2+1/2-1/3+1/3-1/4+...+1/9-1/10)

=1-1/10

=9/10

dễ 

1/2^2=1/1.2

1/3^2=1/2.3

1/4^2=1/3.4

....

1/10^2=1/9.10

1/1.2+1/2.3+1/3.4+...+1/9.10

=(1-1/2+1/2-1/3+1/3-1/4+...+1/9-1/10)

=1-1/10

=9/10

S1 = 1-(1/2*2 + 1/3*3 + 1/4*4 +....+1/10*10)
Coi A = 1/2*2 +1/3*3 +1/4*4 +...+1/10*10
Ta thấy : 1/2*2 < 1/1*2
              1/3*3 < 1/2*3
           ...1/10*10 < 1/9*10
      => A < 1/1*2 + 1/2*3 + 1/3*4 +...+1/9*10 = 9/10
      => 1 - A > 1 - 9/10
       => S1 > 1/10 > 0

Nhận xét:

1/2 mũ 2 + 1/3 mũ 2 + 1/4 mũ 2 + ... + 1/10 mũ 2 <1/1.2 + 1/2.3 + 1/3.4 + ... + 1/ 9.10

Ta có:

A = 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/9.10

A = 1 - 1/2 + 1/2 - 1/3 + 1/3 -1/4 + ... + 1/9 - 1/10

A = 1 - 1/10

A = 9/10 

=> A < 1

Mà S < A

=> S < 1

S  <  1

1/2^2<1/1*2

1/3^2<1/2*3

...

1/10^2<1/9*10

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

=>A<9/10

=>A<1