`S=` $\frac{1}{1.2.3}+$ $\frac{1}{2.3.4}+$ $\frac{1}{3.4.5}+…+$ $\frac{1}{2017.2018.2019}$

Đáp án:

`⇒S=\frac{1018585}{4074342}`

Giải thích các bước giải:

 `S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+….+\frac{1}{2017.2018.2019}`

`⇒S=\frac{1}{2}.(\frac{2}{1.2.3}+\frac{2}{2.3.4}+….+\frac{2}{2017.2018.2019})`

`⇒S=\frac{1}{2}.(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+….+\frac{1}{2017.2018}-\frac{1}{2018.2019})`

`⇒S=\frac{1}{2}.(\frac{1}{2}-\frac{1}{4074342})`

`⇒S=\frac{1018585}{4074342}`