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Đặt \(A=\frac{\frac{1}{2020}+\frac{2}{2019}+\frac{3}{2018}+...+\frac{2019}{2}+\frac{2020}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)
\(A=\frac{1+\left(\frac{1}{2020}+1\right)+\left(\frac{2}{2019}+1\right)+\left(\frac{3}{2018}+1\right)+...+\left(\frac{2019}{2}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)
\(A=\frac{\frac{2021}{2021}+\frac{2021}{2020}+\frac{2021}{2019}+...+\frac{2021}{2}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)
\(A=\frac{2021\left(\frac{1}{2021}+\frac{1}{2020}+\frac{1}{2019}+...+\frac{1}{2}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}=2021\)
Làm theo cách của Trắng nha ,
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{1}{2^2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{1}{4}+\frac{1}{2}-\frac{1}{2019}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{3}{4}-\frac{1}{2019}< \frac{3}{4}\left(Đpcm\right)\)
Ta có: \(\frac{1}{2^2}=\frac{1}{2^2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
...................
\(\frac{1}{2019^2}< \frac{1}{2018.2019}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2019^2}< \frac{1}{2^2}+\frac{1}{2.3}+...+\frac{1}{2018.2019}\)
\(=\frac{1}{2^2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{2019}\)
\(=\frac{1}{4}+\frac{2}{4}-\frac{1}{2019}\)
\(=\frac{3}{4}-\frac{1}{2019}\)\(< \frac{3}{4}\)
\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2019^2}< \frac{3}{4}\)
Điều phải chứng minh
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{2019^2}\)
\(\Rightarrow A=\frac{1}{2^2}+\left(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2019^2}\right)\)
\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2018.2019}\right)\)
\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2}-\frac{1}{2019}\right)\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{2019}=\frac{3}{4}-\frac{1}{2019}< \frac{3}{4}\)
\(\Rightarrow A< \frac{3}{4}\)
\(F=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2020}\)
Trả lời:
\(F=1.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+...+\frac{1}{2019}.\frac{1}{2020}\)
\(F=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(F=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(F=\frac{1}{1}-\frac{1}{2020}\)
\(F=\frac{2019}{2020}\)
\(S=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2017}\)
\(S=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{2035153}\)
\(S=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{4070306}\)
\(S=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{2017.2018}\)
\(S=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2017.2018}\right)\)
\(S=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2018}\right)\)
\(S=2.\left(\frac{1}{2}-\frac{1}{2018}\right)=2.\frac{504}{1009}=\frac{1008}{1009}\)
Vậy \(S=\frac{1008}{1009}\)
1. Bài giải:
Đặt \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1002}\)
\(\Rightarrow\frac{1}{2}A=A-\frac{1}{2}A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1000}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1002}\right)\)
\(\Rightarrow\frac{1}{2}A=1-\frac{1}{1002}=\frac{1001}{1002}\Rightarrow A=\frac{2002}{1002}=\frac{1001}{501}\)
Vậy \(A=\frac{1001}{501}\)
tương tự: https://lazi.vn/edu/exercise/tinh-tong-s-1-1-11-2-1-1-1-2-3-1-1-2-3-2018
\(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2019}\)
= \(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{2039190}\)
= \(2.\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{4078380}\right)\)
= \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\right)\)
= \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\right)\)
= \(2.\left(\frac{1}{2}-\frac{1}{2020}\right)\)
= \(2.\frac{1009}{2020}\)
= \(\frac{1009}{1010}\)