Question

S = \(\dfrac{-1}{1.2}\) - \(\dfrac{1}{2.3}\) - \(\dfrac{1}{3.4}\) - .... - \(\dfrac{1}{\left(n-1\right).n}\) ( tính)

help yourself with home !!!!!!!!!!!!

Answer 1

Theo để ra ta có

\(-S=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right)n}\)

\(\Rightarrow-S=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)

\(\Rightarrow-S=1-\dfrac{1}{n}=\dfrac{n-1}{n}\)

\(\Rightarrow S=\dfrac{1-n}{n}\)

Answer 2

Ta có:

\(S=\dfrac{-1}{1.2}-\dfrac{1}{2.3}-\dfrac{1}{3.4}-...-\dfrac{1}{\left(n-1\right)n}\\ S=-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}\right)\\ S=-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\right)\\ S=-\left(1-\dfrac{1}{n}\right)\\ S=-\dfrac{n-1}{n}\)

Vậy \(S=-\dfrac{n-1}{n}\)