`A=1/2+1/6+1/12+1/20+1/30+...+1/9900`

`=1/(1xx2)+1/(2xx3)+1/(3xx4)+1/(4xx5)+1/(5xx6)+...+1/(99xx100)`

`=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+...+1/99-1/100`

`=1/1-1/100`

`=100/100-1/100`

`=99/100`

$A=1/2+1/6+1/12+1/20+1/30+...+1/9900$

$=1/(1xx2)+1/(2xx3)+1/(3xx4)+1/(4xx5)+1/(5xx6)+...+1/(99xx100)$

$=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+...+1/99-1/100$

$=1/1-1/100$

$=100/100-1/100$

$=99/100$

ta có: 
1/2+1/6+...+1/9900 
=1/1.2+1/2.3...+1/99.100 
=1-1/2+1/2-1/3+1/3-...+1/99-1/100 
=1-1/100 
=99/100

\(A=\frac{1}{2}+\frac{1}{6}+\cdot\cdot\cdot+\frac{1}{9900}\)

\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+\cdot\cdot\cdot+\frac{1}{99\times100}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\frac{1}{100}\)

\(A=\frac{99}{100}\)

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9900}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{1}-\frac{1}{100}\)

\(A=\frac{99}{100}\)

ta có: 

A= 1/6+1/12+1/20+1/30+1/42+1/56

= 1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8

= 1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8

= 1/2-1/8

= 3/8

vậy A= 3/8

TA CÓ: 

A=3/8 ban nha k cho minh voi

\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\)

\(\Rightarrow B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

         \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)

           \(=1-\frac{1}{100}\)

            \(=\frac{99}{100}\)

\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)

\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

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

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

\(B=1-\frac{1}{100}=\frac{99}{100}\)

~ Hok tốt ~

`1/2 + 1/6 + 1/12 +1/20 + 1/30 + 1/42`

`=1/(1.2) + 1/(2.3) + 1/(3.4) + 1/(4.5) + 1/(5.6) + 1/(6.7)`

`=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7`

`=1-1/7`

`=6/7`

\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\)

\(3A=3\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}\right)\)

\(3A=\frac{3}{4}+\frac{3}{12}+\frac{3}{36}+\frac{3}{108}+\frac{3}{324}+\frac{3}{927}\)

\(3A=\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\)

\(2A=3A-A\)

\(2A=\left(\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\right)-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}\right)\)

\(2A=\frac{3}{4}-\frac{1}{927}\)

\(2A=\frac{729-1}{972}=\frac{728}{972}=\frac{182}{243}\)

\(A=\frac{182}{243}:\frac{1}{2}\)

\(A=\frac{364}{243}\)

A=1/4+1/12+1/36+1/108+1/324+1/972

=243/972+81/972+27/972+9/972+3/972+1/972

=364/972

=91/243

\(B=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}\)

\(B=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{30}\right)+\left(1-\frac{1}{42}\right)+\left(1-\frac{1}{56}\right)\)

\(B=1.9-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{56}\right)\)

\(B=9-\left[\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{56}\right)\right]\)

\(B=9-\left(1-\frac{1}{56}\right)\)

\(B=9-\frac{55}{56}\)

\(B=\frac{449}{56}\)