Răspuns:
[tex]\dfrac{1}{n + 1} - \dfrac{1}{n} = \dfrac{n}{n(n + 1)} - \dfrac{n + 1}{n(n + 1)} = \dfrac{n - n - 1}{n(n + 1)} = - \dfrac{1}{n(n + 1)}[/tex]
Reținem și formula:
[tex]\boldsymbol{\dfrac{1}{n} - \dfrac{1}{n + 1} = \dfrac{1}{n(n + 1)}}[/tex]
[tex]S = 1 - \bigg(\dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dfrac{1}{3 \cdot 4} + ... + \dfrac{1}{9 \cdot 10} \bigg) \\[/tex]
[tex]S = 1 - \bigg(\dfrac{1}{1} - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{9} - \dfrac{1}{10} \bigg) \\[/tex]
Se reduc termenii asemenea.
[tex]S = 1 - \bigg(1 - \dfrac{1}{10} \bigg) = 1 - 1 + \dfrac{1}{10} \\[/tex]
[tex]\bf S = \dfrac{1}{10}[/tex]
