B = 1/1.6 + 1/6.11 + 1/11.16 + ... + 1/496.501

B x 5 = 5/1.6 + 5/6.11 + 5/11.16 + ... + 5/496.501

B x 5 = 1 - 1/6 + 1/6 - 1/11 + 1/11 - 1/16 + ... + 1/496 - 1/501

B x 5 = 1 - 1/501

B x 5 = 500/501

      B = 500/501 : 5

       B = 100/501

\(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{46.51}\)

\(=\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{46.51}\right)\)

\(=\frac{1}{5}.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{51}\right)\)

\(=\frac{1}{5}.\left(1-\frac{1}{51}\right)\)

\(=\frac{1}{5}.\left(\frac{51}{51}-\frac{1}{51}\right)\)

\(=\frac{1}{5}.\frac{50}{51}\)

\(=\frac{10}{51}\)

Chúc bạn học tốt !!! 

=1/5.(1-1/51)

=1/5.50/51

=10/51

Đặt \(A=\frac{1}{1\cdot6}+\frac{1}{6\cdot11}+\frac{1}{11\cdot16}+...+\frac{1}{46\cdot51}\)

\(5A=5\left(\frac{1}{1\cdot6}+\frac{1}{6\cdot11}+\frac{1}{11\cdot16}+...+\frac{1}{46\cdot51}\right)\)

\(5A=\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{5}{46\cdot51}\)

\(5A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{46}-\frac{1}{51}\)

\(5A=1-\frac{1}{51}\)

\(5A=\frac{50}{51}\Rightarrow A=\frac{50}{51}:5\Rightarrow A=\frac{10}{51}\)

\(A=\)\(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+\frac{1}{16.21}+...+\frac{1}{51.56}\)

\(5A=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{51.56}\)

\(5A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{51}-\frac{1}{56}\)

\(5A=1-\frac{1}{56}=\frac{55}{56}\)

\(A=\frac{55}{56}\div5=\frac{55}{56}.\frac{1}{5}=\frac{11}{56}\)

\(A=\frac{1}{5}\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{5}{96\cdot101}\right)\)

\(A=\frac{1}{5}\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)\)

\(A=\frac{1}{5}\left(\frac{1}{1}-\frac{1}{101}\right)\)

\(A=\frac{1}{5}\cdot\frac{100}{101}\)

\(A=\frac{20}{101}\)

A = 1/5(1-1/6+1/6-1/11+1/11-1/16+.....+1/96-1/101)

    = 1/5(1-1/101)=20/101

ta có : 1/1.6+1/6.11+1/11.16+....+1/96.101

= 1/5.5/1.6+ 1/5.5/6.11+1/5.5/11.16+...+1/5.5/96.101

=1/5 . ( 5/1.6+5/6.11+5/11.16+...+5/96.101)

=1/5 . ( 1/1-1/6 +1/6-1/11+1/11-1/16+....+1/96-1/101)

=1/5 . (1/1-1/101)

=1/5 . 100/101

= 20/101

5A=\( 1-{1\over 6}+{1\over 6}-{1\over 11}+...{1\over 96}-{1\over 101}\)

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

suy ra A =\({20 \over 101}\)

\(A=\frac{1}{5}\cdot\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+\frac{...5}{96\cdot101}\right)\)

\(A=\frac{1}{5}\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)\)

\(A=\frac{1}{5}\left(\frac{1}{1}+0+0+0+...+0-\frac{1}{101}\right)\)

\(A=\frac{1}{5}\left(1-\frac{1}{101}\right)\)

\(A=\frac{1}{5}\cdot\frac{100}{101}\)

\(A=\frac{20}{101}\)

Nhân A với 5 ta được

A .5=5.(1/1.6 +1/6.11+1/11.16+....+1/496.561)

A .5=5/1.6+5/6.11+5/11.16+...+5/496.561

A .5=1-1/6+1/6-1/11+1/11-1/16+...+1/496-1/561

A .5=1-1/561

A .5=560/561

A    =560/561 : 5 =112/561

\(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{496.561}\)

\(=\left(\frac{1}{1}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{11}\right)+\left(\frac{1}{11}-\frac{1}{16}\right)+...+\left(\frac{1}{496}-\frac{1}{561}\right)\)

\(=\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{496}-\frac{1}{561}\)

\(=\frac{1}{1}-\frac{1}{561}\)

\(=\frac{561}{561}-\frac{1}{561}\)

\(=\frac{560}{561}\)

\(\Rightarrow\) Vậy, \(A=\frac{560}{561}\)

=1-1/6+1/6-1/11+1/11-1/16+...+1/496-1/561

=1-1/561

=560/561

Ta có:

Số hạng thứ nhất: \(\frac{1}{1.6}=\frac{1}{\left(5.1-4\right).\left(5.1+1\right)}\)

Số hạng thứ 2: \(\frac{1}{6.11}=\frac{1}{\left(5.2-4\right).\left(5.2+1\right)}\)

Số hạng thứ 3: \(\frac{1}{11.16}=\frac{1}{\left(5.3-4\right)+\left(5.3+1\right)}\)

.......

Số hạng thứ n = \(\frac{1}{\left(5.n-4\right)+\left(5.n+1\right)}\)

Vậy số hạng 100 của dãy đó là: \(\frac{1}{\left(5.100-4\right).\left(5.100+1\right)}=\frac{1}{496.501}\)