Giá trị của tổng \(\dfrac{1}{{5.10}} + \dfrac{1}{{10.15}} + \dfrac{1}{{15.20}} + ... + \dfrac{1}{{90.95}} + \dfrac{1}{{95.100}}\) là:

Xét: \(B = \dfrac{1}{{5.10}} + \dfrac{1}{{10.15}} + \dfrac{1}{{15.20}} + ... + \dfrac{1}{{90.95}} + \dfrac{1}{{95.100}}\)

Thì \(5B = \dfrac{5}{{5.10}} + \dfrac{5}{{10.15}} + \dfrac{5}{{15.20}} + ... + \dfrac{5}{{90.95}} + \dfrac{5}{{95.100}}\)

Ta có: \(\dfrac{5}{{5.10}} = \dfrac{1}{5} - \dfrac{1}{{10}}\)

\(\dfrac{5}{{10.15}} = \dfrac{1}{{10}} - \dfrac{1}{{15}}\)

\(\dfrac{5}{{15.20}} = \dfrac{1}{{15}} - \dfrac{1}{{20}}\)

…

\(\dfrac{5}{{90.95}} = \dfrac{1}{{90}} - \dfrac{1}{{95}}\)

\(\dfrac{5}{{95.100}} = \dfrac{1}{{95}} - \dfrac{1}{{100}}\)

Suy ra:  \(5B = \dfrac{5}{{5.10}} + \dfrac{5}{{10.15}} + \dfrac{5}{{15.20}} + ... + \dfrac{5}{{90.95}} + \dfrac{5}{{95.100}}\)

\( = \dfrac{1}{5} - \dfrac{1}{{10}} + \dfrac{1}{{10}} - \dfrac{1}{{15}} + .... + \dfrac{1}{{90}} - \dfrac{1}{{95}} + \dfrac{1}{{95}} - \dfrac{1}{{100}}\)

\(\begin{array}{l} = \dfrac{1}{5} - \dfrac{1}{{100}}\\ = \dfrac{{19}}{{100}}\end{array}\)

Do đó: \(B = \dfrac{{19}}{{100}}:5 = \dfrac{{19}}{{500}}\)