a=1+1/3+1/6+1/10+...+1/4950
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\(A=\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{9900}\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=2\cdot\dfrac{49}{100}=\dfrac{98}{100}>\dfrac{1}{4}\)
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 ! ~
\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{4950}\)
\(\Rightarrow2A=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{9900}\)
\(\Rightarrow2A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow2A=\dfrac{1}{2}-\dfrac{1}{100}\)
\(\Rightarrow A=1-\dfrac{1}{50}\)
\(\Rightarrow A=\dfrac{49}{50}\)
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
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
1/3+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
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 = \(\dfrac{99}{50}\)