Question

Cho A=3/1*2*3+3/2*3*4+3/3*4*5+.........+3/2015*2016*2017.

So sánh A với 1.

Answer 1

A=\(\dfrac{3}{1\cdot2\cdot3}+\dfrac{3}{2\cdot3\cdot4}+...+\dfrac{3}{2015\cdot2016\cdot2017}\)

Nhận xét:\(\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}=\dfrac{n+1-n+1}{\left(n-1\right)n\left(n+1\right)}=\dfrac{2}{\left(n-1\right)n\left(n+1\right)}\)

=>A=\(3\cdot\dfrac{1}{2}\cdot\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{2015\cdot2016}-\dfrac{1}{2016\cdot2017}\right)=\dfrac{3}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{2016\cdot2017}\right)=\dfrac{3}{4}-\dfrac{3}{2.2016.2017}< \dfrac{3}{4}< 1\)

Vậy A<1