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

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

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

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

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

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{1}-\frac{1}{100}\)
\(\frac{100-1}{100}\)
\(\frac{99}{100}\)

 

99/100

ta có :\(\frac{1}{1\cdot2}=\frac{1}{1}-\frac{1}{2}\) 

          \(\frac{1}{2\cdot3}=\frac{1}{2}-\frac{1}{3}\)

           \(\frac{1}{3\cdot4}=\frac{1}{3}-\frac{1}{4}\)

            ......

          \(\frac{1}{99\cdot100}=\frac{1}{99}-\frac{1}{100}\)

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

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

B=1/3*4-1/4*5-....-1/99*100

-(1/3*4+1/4*5+...+1/99*100)

=-(1/3-1/4+1/5-1/5+...+1/99-1/100)

=-(1/3-1/100)

=-(100/300-3/300)=-97/300

1/3x4 + 1/4x5 + 1/5x6 +...+ 1/97x 98 + 1/98x99 + x =1

=> 1/3-1/4+1/4-1/5+1/5-1/6+....+1/97-1/98 + 1/98-1/99 +x = 1

=> 1/3 - 1/99 +x=1

=>  32/99+x=1

=> x= 1-32/99

=> x = 67/99

67/99

\(\text{Đặt }A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

\(\Leftrightarrow A=\frac{1}{2}-\frac{1}{100}\)

\(\Leftrightarrow A=\frac{49}{100}\)

1/2x3+1/3x4+....+1/99x100

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

=1-1/100

=99/100

49/100

A = 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + ........... + 1/99 - 1/100

A = 1/4 - 1/100

A = 6/25

=1/4-1/5+1/5-1/6+...+1/99-1/100

=1/4-1/100

=6/25

A = \(\frac{1}{4x5}\)+ \(\frac{1}{5x6}\)+ \(\frac{1}{6x7}\)+..........+ \(\frac{1}{99x100}\)

A = \(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\)

A = \(\frac{1}{4}-\frac{1}{100}\)

A = \(\frac{6}{25}\)

\(A=\frac{1}{2\times3}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{98\times99}\)

\(=\frac{1}{6}+\frac{5-4}{4\times5}+\frac{6-5}{5\times6}+...+\frac{1}{98\times99}\)

\(=\frac{1}{6}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{99}\)

\(=\frac{1}{6}+\frac{1}{4}-\frac{1}{99}=\frac{161}{396}>\frac{160}{400}=\frac{2}{5}\)

\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

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

\(\Rightarrow\frac{1}{2}-0+0+...+0-\frac{1}{100}\)

\(\Rightarrow\frac{50}{100}-\frac{1}{100}=\frac{49}{100}\)

\(A=\frac{1.98+2.97+3.96+...+98.1}{1.2+2.3+3.4+...+98.99}=\frac{1.\left(100-2\right)+2\left(100-3\right)+3\left(100-4\right)+...+98\left(100-99\right)}{1.2+2.3+3.4+...+98.99}\)

\(A=\frac{1.100-1.2+2.100-2.3+3.100-3.4+...+98.100-98.99}{1.2+2.3+3.4+...+98.99}\)

\(A=\frac{\left(1.100+2.100+3.100+...+98.100\right)-\left(1.2+2.3+3.4+...+98.99\right)}{1.2+2.3+3.4+...+98.99}\)

\(A=\frac{100\left(1+2+3+...+98\right)}{1.2+2.3+3.4+...+98.99}-1\)

Ta có: 1+2+3+...+98=98.99:2=4851

Đặt B=1.2+2.3+3.4+...+98.99  => 3B=1.2.3+2.3.3+3.4.3+...+98.99.3 = 1.2.3+2.3.(4-1)+3.4(5-2)+...+98.99(100-97)

=> 3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+98.99.100-97.98.99 = 98.99.100

=> B=33.98.100. Thay vào A được:

\(A=\frac{100.4851}{33.98.100}-1=\frac{3}{2}-1=\frac{1}{2}\)