`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`

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

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

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

$=1/1-1/100$

$=100/100-1/100$

$=99/100$

1/2+1/6+1/12+...+1/9900
=1/(1*2)+1/(2*3)+1/(3*4)+...+1/(99*100)
=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
=1-1/100

=99/100

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

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

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

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9900}\)
\(=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)

\(B=\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)\(=\left(1-\frac{1}{100}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+...\left(\frac{1}{99}-\frac{1}{99}\right)=\left(\frac{100}{100}-\frac{1}{100}\right)+0+...+0=\frac{99}{100}\)Vậy B=99/100

MK k chắc nữa

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

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

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

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

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

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

Giải 

\(A=1+2+3+4+5+...+99+100\)

Số số hạng của A là: \(\left(100-1\right)\div1+1=100\)(số hạng)

Tổng A là: \(\frac{\left(100+1\right)\times100}{2}=5050\)

Vây A=5050

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

\(B=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)

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

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

Vậy \(B=\frac{99}{100}\)

minh cam thay de hoi sai

A = 1 + 2 + 3 + ... + 99 + 100

A = 100 + 99 + ... + 2 + 1

2A = 101 + 101 +... + 101 + 101 ( 100 số hạng )

A = 101 . 100 : 2 = 5050

Vậy A = 5050 

B = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + ... + 1/9900

B = 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/99.100

B = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100

B = 1/1 - 1/100

B = 99/100

Vậy B = 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

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

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

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

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

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

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

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

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

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

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

T= 1 - 1/2 + 1/2 - 1/3 + ......+ 1/99 - 1/100

  = 1 - 1/100

  = 99/100

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

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

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

Vậy \(t=\frac{99}{100}\)

dấu . hiểu là phép nhân nhé

tớ cũng nghĩ là 99/100