Cho \(A = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{{2022}}\)
và \(B = \frac{{2021}}{1} + \frac{{2020}}{2} + \frac{{2019}}{3} + ... + \frac{1}{{2021}}\)
Tính tỉ số \(\frac{B}{A}\).
Tách \(2021\) thành \(2020 +1\) và biến đổi biểu thức \(B = 2020\left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2021}} + \frac{1}{{2022}}} \right)\)
Ta có:
\(A = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{{2022}}\)
\(\begin{array}{l}B = \frac{{2021}}{1} + \frac{{2020}}{2} + \frac{{2019}}{3} + ... + \frac{1}{{2021}}\\ = 2021 + \frac{{2020}}{2} + \frac{{2019}}{3} + ... + \frac{1}{{2021}}\\ = 2020 + \frac{{2020}}{2} + \frac{{2019}}{3} + ... + \frac{1}{{2021}} + 1\\ = \left( {1 + \frac{{2020}}{2}} \right) + \left( {1 + \frac{{2019}}{3}} \right) + ... + \left( {1 + \frac{1}{{2021}}} \right) + 1\\ = \frac{{2022}}{2} + \frac{{2022}}{3} + ... + \frac{{2022}}{{2021}} + \frac{{2022}}{{2022}}\\ = 2022\left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2021}} + \frac{1}{{2022}}} \right)\end{array}\)
Khi đó: \(\frac{B}{A} = \frac{{2022\left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2021}} + \frac{1}{{2022}}} \right)}}{{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{{2022}}}} = 2022\)
Vậy tỉ số: \(\frac{B}{A} = 2022\)







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