HL
1
K
Khách
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LQ
5
1 tháng 8 2017
A = \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{4950}\)
A = \(2.\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{9900}\right)\)
A = \(2.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
A = \(2.\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)
A = \(1-\dfrac{1}{50}\)
A = \(\dfrac{49}{50}\)
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PT
1 tháng 8 2017
\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{4950}\)
\(\Rightarrow2A=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{9900}\)
\(\Rightarrow2A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow2A=\dfrac{1}{2}-\dfrac{1}{100}\)
\(\Rightarrow A=1-\dfrac{1}{50}\)
\(\Rightarrow A=\dfrac{49}{50}\)
HT
1
NT
1
NL
Nguyễn Lê Phước Thịnh
CTVHS
3 tháng 7 2022
\(A=\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{9900}\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=2\cdot\dfrac{49}{100}=\dfrac{98}{100}>\dfrac{1}{4}\)
XO
31 tháng 8 2020
\(1+\frac{1}{2}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{4950}=2\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\right)\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2\left(1-\frac{1}{100}\right)=2.\frac{99}{100}=\frac{99}{50}\)
NT
1
NT
0
Ta có công thức tổng quát sau:
\(1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}\)
\(=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)
Ta có: \(\left(1-\frac13\right)\left(1-\frac16\right)\cdot\ldots\cdot\left(1-\frac{1}{4950}\right)\)
\(=\left(1-\frac26\right)\left(1-\frac{2}{12}\right)\cdot\ldots\cdot\left(1-\frac{2}{9900}\right)\)
\(=\left(1-\frac{2}{2\cdot3}\right)\left(1-\frac{2}{3\cdot4}\right)\cdot...\cdot\left(1-\frac{2}{99\cdot100}\right)\)
\(=\frac{\left(2+2\right)\left(2-1\right)}{2\left(2+1\right)}\cdot\frac{\left(3+2\right)\left(3-1\right)}{3\left(3+1\right)}\cdot\ldots\cdot\frac{\left(99+2\right)\left(99-1\right)}{99\left(99+1\right)}\)
\(=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\ldots\cdot\frac{101\cdot98}{99\cdot100}\)
\(=\frac{4\cdot5\cdot\ldots\cdot101}{3\cdot4\cdot\ldots\cdot100}\cdot\frac{1\cdot2\cdot\ldots\cdot98}{2\cdot3\cdot\ldots\cdot99}=\frac{101}{3}\cdot\frac{1}{99}=\frac{101}{297}\)