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