\(x+\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+\left(x+\frac{1}{12}\right)+...+\left(x+\frac{1}{9900}\right)=2\)

=> \(x+\left(x+\frac{1}{1.2}\right)+\left(x+\frac{1}{2.3}\right)+\left(x+\frac{1}{3.4}\right)+...+\left(x+\frac{1}{99.100}\right)=2\)

 => \(\left(x+x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)=2\)(100 hạng tử x)

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

=>  \(100x+1-\frac{1}{100}=2\)

=> \(100x+\frac{99}{100}=2\)

=> \(100x=\frac{101}{100}\)

=> \(x=\frac{101}{10000}\)

Đặt A=1+2+2²+2³+...+2¹⁰⁰

suy ra 2A=2+2²+2³+...+2¹⁰⁰

suy ra 2A-A=(2+2²+2³+...+2¹⁰¹)-(1+2+2²+2³+...+2¹⁰⁰⁾

                 =2¹⁰¹-1

Vậy 1+2+2²+2³+...+2¹⁰⁰=2¹⁰¹-1

 A=1+2+2^2+2^3+...+2^100

2A=2+2²+2³+2⁴+...+2¹⁰¹

2A-A=2¹⁰¹-1

Vậy A= 2¹⁰¹-1

Ta có: \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}+\frac{1}{10100}\)

     \(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}+\frac{1}{100.101}\)

     \(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}\)

     \(=1-\frac{1}{101}\)

     \(=\frac{100}{101}\)

Tương đương \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}+\frac{1}{100.101}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)

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

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

= 1/1 - 1/100

= 100/100 - 1/100

= 99/100

1/2+1/6+1/12+1/20+...+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

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

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

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

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

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

 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

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

MK k chắc nữa

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

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

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

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

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

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