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a) \(\dfrac{1}{2!}+\dfrac{2}{3!}+...+\dfrac{2018}{2019!}\\ =\left(\dfrac{1}{1!}-\dfrac{1}{2!}\right)+\left(\dfrac{1}{2!}-\dfrac{1}{3!}\right)+...+\left(\dfrac{1}{2018!}-\dfrac{1}{2019!}\right)\\ =1-\dfrac{1}{2019!}< 1\)
b) \(\dfrac{1\cdot2-1}{2!}+\dfrac{2\cdot3-1}{3!}+...+\dfrac{999\cdot1000-1}{1000!}\\ =\dfrac{1\cdot2}{2!}-\dfrac{1}{2!}+\dfrac{2\cdot3}{3!}-\dfrac{1}{3!}+...+\dfrac{999-1000}{1000!}-\dfrac{1}{1000!}\\ =\dfrac{1}{1!}-\dfrac{1}{2!}+\dfrac{1}{1!}-\dfrac{1}{3!}+\dfrac{1}{2!}-\dfrac{1}{4!}+...+\dfrac{1}{999!}+\dfrac{1}{1000!}\\ =1+1-\dfrac{1}{1000!}\\ =2-\dfrac{1}{1000!}< 2\)
Ta có:
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=1-\dfrac{1}{50}\)
\(\Rightarrow A=\dfrac{49}{50}\)
Vậy \(A=\dfrac{49}{50}.\)
1) Ta có :
\(\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{n+1}{n\left(n+1\right)}-\dfrac{n}{n\left(n+1\right)}=\dfrac{1}{n\left(n+1\right)}\)
Vậy \(\dfrac{1}{n\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\rightarrowđpcm\)
2) \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+............+\dfrac{1}{99.100}\)
\(\Leftrightarrow A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+......+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Leftrightarrow A=1-\dfrac{1}{100}\)
\(\Leftrightarrow A=\dfrac{99}{100}\)
Bài 1:
a) Ta có: \(\dfrac{7^4\cdot3-7^3}{7^4\cdot6-7^3\cdot2}\)
\(=\dfrac{7^3\cdot\left(7\cdot3-1\right)}{7^3\cdot2\left(7\cdot3-1\right)}\)
\(=\dfrac{1}{2}\)
c) Ta có: \(E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)
\(\Leftrightarrow\dfrac{1}{3}\cdot E=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\)
\(\Leftrightarrow E-\dfrac{1}{3}\cdot E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\right)\)
\(\Leftrightarrow E\cdot\dfrac{2}{3}=1-\dfrac{1}{3^{101}}\)
\(\Leftrightarrow E=\dfrac{3-\dfrac{3}{3^{101}}}{2}=\dfrac{1-\dfrac{1}{3^{100}}}{2}\)
999/1000(hình như v)
Áp dụng công thức \(\dfrac{1}{k\left(k+1\right)}=\dfrac{1}{k}-\dfrac{1}{k+1}\), ta có:
\(A=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{999}-\dfrac{1}{1000}\right)=1-\dfrac{1}{1000}=\dfrac{999}{1000}\)