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a,A=2^0+2^1+2^2+...+2^2014
2A=2^1+2^2+2^3+...+2^2015
2A-A=(2^1+2^2+2^3+...+2^2015)-(2^0+2^1+2^2+...+2^2014)
A=2^2015-2^0=2^2015-1=B
=>A=B
b,A=2014.2016=2014.(2015+1)=2014.2015+2014
B=2015^2=2015.2015=(2014+1).2015=2014.2015+2015
Vì 2014<2015 => A<B.
\(\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)\cdot..\cdot\left(\frac{1}{10^2}-1\right)\)
\(=\left(\frac{1}{2}\cdot\frac{1}{2}-1\right)\cdot\left(\frac{1}{3}\cdot\frac{1}{3}-1\right)\cdot...\cdot\left(\frac{1}{10}\cdot\frac{1}{10}-1\right)\)
\(=\left(\frac{1}{4}-1\right)\cdot\left(\frac{1}{9}-1\right)\cdot...\cdot\left(\frac{1}{100}-1\right)\)
\(=\frac{-3}{4}\cdot\frac{-8}{9}\cdot...\cdot\frac{-99}{100}\)
\(=\frac{\left(-1\right).\left(-3\right)}{2.2}\cdot\frac{\left(-2\right).\left(-4\right)}{3.3}\cdot...\cdot\frac{\left(-9\right).\left(-11\right)}{10.10}\)
\(=\frac{\left(-1\right).\left(-2\right)....\left(-9\right)}{2.3....10}\cdot\frac{\left(-3\right).\left(-4\right)....\left(-11\right)}{2.3.....10}\)
\(=\frac{-1}{10}\cdot\frac{-11}{2}=\frac{-11}{20}\)
B=2^2015=2^2014.2=2^2014+2^2014
=2^2014+2^2013.2=2^2014+2^2013+2^2013
=2^2014+2^2013+...+2^3.2=2^2014+2^2013+...+2^3+2^3
=2^2014+2^2013+...+2^3+2^2.2=2^2014+2^2013+...+2^3+2^2+2^2
=2^2014+2^2013+...2^3+2^2+2.2=2^2014+2^2013+...+2^3+2^2+2+2
A=1+2+2^2+2^3+...+2^2013+2^2014
=>B> A
2.A = 2.(1+2+22+...+22014)=2+22+23+...22015
2A-A=A=(2+22+...+22015)-(1+2+22+...+22014)
=A=22015-1va B=22015
=A<B
a/M=2/3.5+2/5.7+2/7.9+.....+2/97.99
M=1/3-1/5+1/5-1/7+..+1/97-1/99
M=1/3-1/99
M=32/99
b)ta có 1/2.3+1/3.4+1/4.5+..+1/2015.2016+1/2016.2017<A
=>1/2-1/3+1/3-1/4+1/4-1/5+..+1/2015-1/2016+1/2016-1/2017<a
1/2-1/2017<A
2/15/4034<A (1)
Ta có
1/1.2+1/2.3+1/3.4+1/4.5+..+1/2015.2016>A
=>1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+..+1/2015-1/2016>A
1-1/2016
2015/2016>A (2)
Từ (1) và (2)=>A không phải là số tự nhiên(đpcm)
A=1+2+22+23+.....+22014
=>2A=2+22+23+.....+22015
=>2A-A=22015-1
=>A=22015-1
=>B-A=22015-(22015-1)=1