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1) Tính C
\(C=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+....+\frac{n-1}{n!}\)
\(=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{n-1}{n!}\)
\(=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{\left(n-1\right)!}-\frac{1}{n!}\)
\(=1-\frac{1}{n!}\)
3) a) Ta có : \(P=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{100}\)
\(=\frac{1}{101}+\frac{1}{102}+....+\frac{1}{199}+\frac{1}{200}\left(đpcm\right)\)
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\(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2004^2}\)
\(=1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2004^2}\right)>1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2003.2004}\right)\)
\(>1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2003}-\frac{1}{2004}\right)\)
\(>1-\left(1-\frac{1}{2004}\right)\)
\(>1-1+\frac{1}{2004}\)
\(>\frac{1}{2004}\left(đpcm\right)\)
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Bạn tham khảo cách làm ở đây: https://olm.vn/hoi-dap/question/528628.html
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Đặt \(A=1-\frac{1}{2^2}-\frac{1}{3^2}-.........-\frac{1}{2020^2}\)
Ta có: \(2^2=2.2< 2.3\)\(\Rightarrow\frac{1}{2.2}>\frac{1}{2.3}\)\(\Rightarrow\frac{1}{2^2}>\frac{1}{2.3}\)
Tương tự, ta có: \(\frac{1}{3^2}>\frac{1}{3.4}\), ........... , \(\frac{1}{2020^2}>\frac{1}{2020.2021}\)
\(\Rightarrow A>1-\frac{1}{2.3}-\frac{1}{3.4}-...........-\frac{1}{2020.2021}\)
\(=1-\left(\frac{1}{2}-\frac{1}{3}\right)-\left(\frac{1}{3}-\frac{1}{4}\right)-.......-\left(\frac{1}{2020}-\frac{1}{2021}\right)\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-..........-\frac{1}{2020}+\frac{1}{2021}\)
\(=1-\frac{1}{2}+\frac{1}{2021}\)\(=\frac{1}{2}+\frac{1}{2021}=\frac{2023}{4042}>\frac{1}{2020}\)
\(\Rightarrow A>\frac{1}{2020}\)
Vì\(\frac{121212}{242424}=\frac{12.20202}{24.20202}=\frac{1}{2}\)nên\(\frac{121212}{242424}=\frac{1}{2}\).
Ta có : \(\frac{121212}{242424}\)= \(\frac{1.121212}{2.121212}\)=\(\frac{1}{2}\)