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 = \(\dfrac{99}{50}\)

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

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

Đáp án là 49/50 nha hatsune miku 

Chúc bạn học tốt!

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 ! ~

\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{4950}\)

\(\Rightarrow2A=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{9900}\)

\(\Rightarrow2A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Rightarrow2A=\dfrac{1}{2}-\dfrac{1}{100}\)

\(\Rightarrow A=1-\dfrac{1}{50}\)

\(\Rightarrow A=\dfrac{49}{50}\)

\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{4950}\)

\(\dfrac{1}{2}A=\dfrac{1}{2}\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{4950}\right)\)

\(\dfrac{1}{2}A=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{9900}\)

\(\dfrac{1}{2}A=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)

\(\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{100}\)

\(\dfrac{1}{2}A=\dfrac{49}{100}\)

\(A=\dfrac{49}{50}\)

x3x

\(A=\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{9900}\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=2\cdot\dfrac{49}{100}=\dfrac{98}{100}>\dfrac{1}{4}\)

A=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

 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