Question

cho A= 1011(1 +1/3+1/5+....+1/2019) và B = 1010(1/2+1/4+1/6+...+1/2020)

so sanh A và B

 

Answer 1

Ta có: \(A-B\)

\(=1011\left(1+\frac13+\frac15+\cdots+\frac{1}{2019}\right)-1010\left(\frac12+\frac14+\cdots+\frac{1}{2020}\right)\)

\(=1010\left(1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2019}-\frac{1}{2020}\right)+\left(1+\frac13+\cdots+\frac{1}{2019}\right)\)

Vì \(1-\frac12>0;\frac13-\frac14>0;\ldots;\frac{1}{2019}-\frac{1}{2020}>0\)

nên \(1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2019}-\frac{1}{2020}>0\)

=>\(1010\left(1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2019}-\frac{1}{2020}\right)>0\)

=>A-B>0

=>A>B