Tính tỉ số \(\frac{A}{B}\) biết

\(A = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ..... + \frac{1}{{2017}} + \frac{1}{{2018}} + \frac{1}{{2019}}\)

\(B = \frac{{2018}}{1} + \frac{{2017}}{2} + \frac{{2016}}{3} + ... + \frac{2}{{2017}} + \frac{1}{{2018}}\)

Lời giải:

Ta có \(B = \frac{{2018}}{1} + \frac{{2017}}{2} + \frac{{2016}}{3} + ... + \frac{2}{{2017}} + \frac{1}{{2018}}\)

\( = 1 + \left( {\frac{{2017}}{2} + 1} \right) + \left( {\frac{{2016}}{3} + 1} \right) + .... + \left( {\frac{1}{{2018}} + 1} \right)\)

\( = \frac{{2019}}{{2019}} + \frac{{2019}}{2} + \frac{{2019}}{3} + .... + \frac{{2019}}{{2018}}\)

= 2019 \(\left( {\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ..... + \frac{1}{{2017}} + \frac{1}{{2018}} + \frac{1}{{2019}}} \right)\).

Ta có \(\frac{A}{B}\)=\(\frac{{\left( {\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ..... + \frac{1}{{2017}} + \frac{1}{{2018}} + \frac{1}{{2019}}} \right)}}{{2019\left( {\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ..... + \frac{1}{{2017}} + \frac{1}{{2018}} + \frac{1}{{2019}}} \right)}} = \frac{1}{{2019}}\).