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\(A=1+\frac{1}{2}+...+\frac{1}{2^{100}}\)
=>\(2A=2+1+\frac{1}{2}+...+\frac{1}{2^{99}}\)
=>2A-A=\(\left(2+1+\frac{1}{2}+...+\frac{1}{2^{99}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{100}}\right)=2-\frac{1}{2^{100}}
=> \(\frac{1}{2}\)A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{101}}\)
=> A - \(\frac{1}{2}\) A = \(\frac{1}{2}\)A = \(\frac{1}{2^{101}}-1\)
=> A = \(\frac{\frac{1}{2^{101}}-1}{2}=\frac{\frac{1}{2^{101}}}{2}-\frac{1}{2}=\frac{1}{2^{102}}-\frac{1}{2}
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= 1/2.2 + 1/3.3 + ... + 1/2018.2018
= ( 1/2 - 1/2) + (1/3 - 1/3) + ... + ( 1/2018 - 1/2018 )
= 0+0+0+0+...+0
=0
75% = 7,5
7,5 > 0 ==>
A<B
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B = 75% => B = 3/4
Ta có :\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}=1-\frac{1}{2018}\)
Vì \(\frac{1}{2018}< \frac{1}{4}\Rightarrow1-\frac{1}{2018}>1-\frac{1}{4}\Rightarrow A>\frac{3}{4}\)=> A > B
\(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2018^2}\)
\(B=75\%=\frac{3}{4}\)
Ta có:\(A=.......\)
\(=\frac{1}{4}+\left(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\right)< \frac{1}{4}+\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\right)\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{2018}=\frac{3}{4}-\frac{1}{2018}< \frac{3}{4}\)
\(\Rightarrow A< B\)
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\(\Rightarrow2A=2+2^2+2^3+...+2^{2003}\\ \Rightarrow2A-A=2+2^2+2^3+...+2^{2003}-1-2-...-2^{2002}\\ \Rightarrow A=2^{2003}-1=B\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\)
\(2A=2+1+\frac{1}{2}+...+\frac{1}{2^{99}}\)
\(2A-A=\frac{1}{2^{100}}-2\) HAY \(A=\frac{1}{2^{100}}-2< 2\)