Lời giải:
\(5A=\frac{6-1}{1.6}+\frac{11-6}{6.11}+\frac{16-11}{11.16}+....+\frac{501-496}{496.501}\)

\(=\frac{6}{1.6}-\frac{1}{1.6}+\frac{11}{6.11}-\frac{6}{6.11}+\frac{16}{11.16}-\frac{11}{11.16}+...+\frac{501}{496.501}-\frac{496}{496.501}\)

\(=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+....+\frac{1}{496}-\frac{1}{501}=1-\frac{1}{501}=\frac{500}{501}\)

$\Rightarrow A=\frac{100}{501}$

\(A=\dfrac{1}{5}\left(\dfrac{1}{1.6}+...+\dfrac{1}{496.501}\right)\)

\(A=\dfrac{1}{5}\left(1-\dfrac{1}{6}+\cdot\cdot\cdot+\dfrac{1}{495}-\dfrac{1}{501}\right)\)

\(A=\dfrac{1}{5}\left(1-\dfrac{1}{501}\right)\)

\(A=\dfrac{1}{5}\cdot\dfrac{500}{501}=\dfrac{100}{501}\)

5S=5.(1/1.6+1/6.11+...+1/496.501)

5S=5/1.6+5/6.11+...+5/496.501

5S=1/1-1/6+1/6-1/11+...+1/496-1/501

5S=1-1/501

5S=500/501

S=500/501:5=100/501

k nhé

ta co:5S=5/1.6+5/6.11+5/11.16+...+5/496.501

             =1-1/6+1/6-1/11+1/11-1/16+.....+1/496-1/501

             =1-1/501=500/501

       =>S=500/501:5=100/501

MK đau tien nha bn

1/1.6 + 1/6.11+ 1/11.16+ ....  

số thứ 100 có dạng 1/(496.501)  

do đó tổng trên bằng :

1/5( 1/1- 1/501)

= 100/ 501

1/1-1/6+1/6-1/11+...+1/496-1/501

=1/1-1/501=500/501

\(A=\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+....+\frac{1}{496}-\frac{1}{501}\right):5\)

\(A=\left(1-\frac{1}{501}\right):5\)

\(A=\frac{500}{501}:5=\frac{100}{501}\)

Ta có : \(A=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{496.501}\)

    \(\Rightarrow\)  \(A=\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{496}-\frac{1}{501}\right) \)

     \(\Rightarrow\)  \(A=\frac{1}{5}\left(1-\frac{1}{501}\right)\)

      \(\Rightarrow\)  \(A=\frac{1}{5}.\frac{501-1}{501}=\frac{1}{5}.\frac{500}{501}\)

       \(\Rightarrow\)  \(A=\frac{1.500}{5.501}=\frac{20}{1.501}=\frac{20}{501}\)     

                                               Vậy   \(A=\frac{20}{501}\)

mk nhầm 1 chút : \(A=\frac{1.100}{5.101}=\frac{100}{1.101}=\frac{100}{101}\)

                         Vậy   \(A=\frac{100}{101}\) chứ ko phải bằng  \(\frac{20}{101}\)  đâu nhé mong bn thông cảm!!!!

A=999/1000

B= ...........

C=..................

\(B=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....++\frac{1}{9}-\frac{1}{10}\)

\(B=1-\frac{1}{10}=\frac{9}{10}\)

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

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

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

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

\(D=\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{496.501}\)

\(D=\frac{1}{5}\cdot\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+.....+\frac{1}{496}-\frac{1}{501}\right)\)

\(D=\frac{1}{5}\cdot\left(1-\frac{1}{501}\right)=\frac{1}{5}\cdot\frac{500}{501}=\frac{100}{501}\)

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^n}\)

 =    \(\frac{2-1}{2}+\frac{2-1}{2^2}+\frac{2-1}{2^3}+...+\frac{2-1}{2^n}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{2^3}+...+\frac{1}{2^{n-1}}-\frac{1}{2^n}\)

 \(=1-\frac{1}{2^n}\)

   =\(\frac{2^n-1}{2^n}\)