a)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}=\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+...+\left(\frac{1}{99}-\frac{1}{101}\right)\)

                                                               \(=1-\frac{1}{101}=\frac{100}{101}\)

b) \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}=\frac{2}{1.3}.\frac{5}{2}+\frac{2}{3.5}.\frac{5}{2}+\frac{2}{5.7}.\frac{5}{2}+...+\frac{2}{99.101}.\frac{5}{2}\)

                                                                \(=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)

                                                                \(=\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)

a)100/101

b)250/101

A)100/101

B)250/101

a.2/1.3+2/3.5+2/5.7+................+2/99.101

1-1/3+1/3-1/5+1/5-1/7+....+1/99-1/101

1-1/101

100/101

b.5/1.3+5/3.5+5/5.7+............+5/99.101

5.2/1.3.2+5.2/3.5.2+5.2/5.7.2+........+5.2+99.101.2

5/2(2/1.3+2/3.5+2/5.7+........+2/99.101)

5/2(1-1/3+1/3-1/5+1/5-1/7+........+1/99-1/101)

5/2(1-1/101)

5/2.100/101

250/101

a) =1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101

    =1-1/101

    =100/101

b) =(2/1.3+2/3.5+2/5.7+...+2/99.101).2,5

    =(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101).2,5

    =(1-1/101).2,5

    =100/101.2,5

    =250/101

c) =(2/2.4+2/4.6+2/6.8+...+2/2008-2/2010).2

    =(1/2-1/4+1/4-1/6+1/6-1/8+...+1/2008-1/2010).2

    =(1/2-1/2010).2

    =1004/1005

\(A=1\times3+3\times5+5\times7+...+99\times101\)

\(=1\left(1+2\right)+3\left(3+2\right)+5\left(5+2\right)+...+99\left(99+2\right)\)

\(=\left(1^2+3^2+5^2+...+99^2\right)+2\left(1+3+5+...+99\right)\)

Ta có:

\(1^2+2^2+3^2+...+n^2=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}\)

⇒ \(A=\left(1^2+2^2+3^2+...+100^2\right)-2^2\left(1^2+2^2+3^2+...+50^2\right)+2\left(1+3+5+...+99\right)\)

\(=\dfrac{100.101.201}{6}+\dfrac{4.50.51.101}{6}+\dfrac{\left(99+1\right).\left[\left(99-1\right):2+1\right]}{2}\)

\(=338350-171700+5000\)

\(=166650+5000=171650\)

=99.101.103

Đặt A = 1.3+3.5+5.7+ ...+ 99.101

ta có : 6A = 1.3.6+3.5.6+5.7.6 +...+99.101.6

               = 1.3.(5+1) + 3.5.( 7-1) +5.7.(9-3) +...+99.101.(103-97)

               =1.3.5+1.3+3.5.7-1.3.5+5.7.9-3.5.7+...+99.101.103-97.99.101

               =1.3+99.101.103

              A =1029900 :6 = 171650   

Answer:

Ta đặt \(A=1.3+3.5+5.7+...+99.101\)

\(\Rightarrow6A=1.3.\left(5+1\right)+3.5.\left(7-1\right)+5.7.\left(9-3\right)+...+99.101.\left(103-97\right)\)

\(\Rightarrow6A=1.3.5+1.3+3.5.7-1.3.5+5.7.9-3.5.7+...+99.101.103-97.99.101\)

\(\Rightarrow6A=3+99.101.103\)

\(\Rightarrow6A=1029900\)

\(\Rightarrow A=1029900:6=171650\)

chuẩn quá

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\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}=\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)

\(\dfrac{2}{3.5}=\dfrac{1}{3}-\dfrac{1}{5}\)

\(\dfrac{2}{5.7}=\dfrac{1}{5}-\dfrac{1}{7}\)

\(\dfrac{2}{99.101}=\dfrac{1}{99}-\dfrac{1}{101}\)

\(\Rightarrow=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

\(=\dfrac{1}{3}-\dfrac{1}{101}\)

\(=\dfrac{98}{303}\)

\(\dfrac{2}{3.5} + \dfrac{2}{5.7} + ..... + \dfrac{2}{99.101}\) 

\(= \dfrac{1}{3} - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{7} + ........ + \dfrac{1}{99} - \dfrac{1}{101}\)

\(= \dfrac{1}{3} - ( \dfrac{1}{5} - \dfrac{1}{5} ) + ........ + ( \dfrac{1}{99} -\dfrac{1}{99} ) - \dfrac{1}{101} \) 

\(= \dfrac{1}{3} - \dfrac{1}{101}\)

\(= \dfrac{101}{303} - \dfrac{3}{303}\)

\(= \dfrac{98}{303}\)