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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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\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{4950}\)
\(\Rightarrow2A=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{9900}\)
\(\Rightarrow2A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow2A=\dfrac{1}{2}-\dfrac{1}{100}\)
\(\Rightarrow A=1-\dfrac{1}{50}\)
\(\Rightarrow A=\dfrac{49}{50}\)
\(A=\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{9900}\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=2\cdot\dfrac{49}{100}=\dfrac{98}{100}>\dfrac{1}{4}\)
F=1+3+10+.....+4851+4950
2F=2+6+12+20+.....+9702+9900
2F=1.2+2.3+3.4+4.5+.........+98.99+99.100
xét A=1.2+2.3+3.4+4.5+.....+98.99+99.100
3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+..+99.100.101-98.99.100
3A=99.100.101
A=333300
Thay A vào F ta có:
2F=333300
F=333300:2=166650
TICK NHA nOBITA kUN
mới học
2F = 2+6+12+20+...+9702+9900 = 1.2+2.3+3.4+4.5+...+99.100
2F = 33.100.101
F=33.50.101=166650
A=1+3+6+10+...+4851+4950
2A=2+6+12+20+...+9702+9900
2A=1.2+2.3+3.4+4.5+...+98.99+99.100
Xét B=1.2+2.3+3.4+4.5+...+98.99+99.100
3B=1.2.3+2.3(4−1)+3.4(5−2)+...+99.100(101−98)
3B=1.2.3+2.3.4−1.2.3+3.4.5−2.3.4+...+99.100.101−98.99.100
3B=99.100.101
B=333300
Thay B vào A ta được:
2A=333300
A=166650
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{4950}=2\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\right)\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2\left(1-\frac{1}{100}\right)=2.\frac{99}{100}=\frac{99}{50}\)
\(\frac{1}{1}\)+ \(\frac{1}{3}\)+ \(\frac{1}{6}\)+ \(\frac{1}{10}\)+.........+\(\frac{1}{4950}\)
<=> 2 . \((\)\(\frac{1}{2}\)+ \(\frac{1}{6}\)+ \(\frac{1}{12}\)+ \(\frac{1}{20}\)+ ........ + \(\frac{1}{9900}\)\()\)
<=> 2 . \((\)\(\frac{1}{1}\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)-.........+ \(\frac{1}{99}\)- \(\frac{1}{100}\)\()\)
<=> 2 . \((\)\(\frac{1}{1}\)- \(\frac{1}{100}\))
<=> 2 . \(\frac{99}{100}\)
<=> \(\frac{198}{100}\)= \(\frac{99}{50}\)