(1-1/3).(1-1/6).(1-1/10).....(1-1/4950)
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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
25 tháng 3 2023
A = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + .....+ \(\dfrac{1}{4950}\)
A = \(\dfrac{2}{2}\) \(\times\) ( 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\)+.......+ \(\dfrac{1}{4950}\))
A = 2 \(\times\) ( \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\)+......+ \(\dfrac{1}{9900}\))
A = 2 \(\times\) ( \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\)+ \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\)+....+ \(\dfrac{1}{99.100}\))
A = 2 \(\times\) ( \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +....+ \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\))
A = 2 \(\times\) ( 1 - \(\dfrac{1}{100}\))
A = 2 \(\times\) \(\dfrac{99}{100}\)
A = \(\dfrac{99}{50}\)
NP
1
13 tháng 8 2015
D=1/3+1/6+1/10+1/15+......+1/4950
=2x（1/6+1/12+1/20+1/30+……+1/9900）
=2x（1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+……+1/99-1/100）
=2x（1/2-1/100）
=1-1/50
=49/50
**** nhé
FZ
2
13 tháng 8 2015
1/3+1/6+1/10+1/15+......+1/4950
=2x（1/6+1/12+1/20+1/30+……+1/9900）
=2x（1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+……+1/99-1/100）
=2x（1/2-1/100）
=1-1/50
=49/50
**** nhé
HT
1
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 = \(\dfrac{49}{50}\)
~ Chúc bạn học giỏi ! ~
PT
1 tháng 8 2017
\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{4950}\)
\(\Rightarrow2A=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{9900}\)
\(\Rightarrow2A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow2A=\dfrac{1}{2}-\dfrac{1}{100}\)
\(\Rightarrow A=1-\dfrac{1}{50}\)
\(\Rightarrow A=\dfrac{49}{50}\)
NT
1
3 tháng 7 2022
\(A=\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{9900}\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=2\cdot\dfrac{49}{100}=\dfrac{98}{100}>\dfrac{1}{4}\)
NT
1
XO
31 tháng 8 2020
\(1+\frac{1}{2}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{4950}=2\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\right)\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2\left(1-\frac{1}{100}\right)=2.\frac{99}{100}=\frac{99}{50}\)