Question

Chứng minh rằng 

\(\dfrac{1}{26.2^3-4^3-0^3}+\dfrac{1}{26.3^3-5^3-1^3}+\dfrac{1}{26.4^3-6^3-2^3}+...+\dfrac{1}{26.2020^3-2022^3-2018^3}< \dfrac{1}{96}\)

Answer 1

\(26n^3-\left(n+2\right)^3-\left(n-2\right)^3=24n^3-24n=24n\left(n-1\right)\left(n+1\right)\)

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

Do đó:

\(VT=\dfrac{1}{24}\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2019.2020.2021}\right)\)

\(=\dfrac{1}{48}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{2019.2020}-\dfrac{1}{2020.2021}\right)\)

\(=\dfrac{1}{48}\left(\dfrac{1}{2}-\dfrac{1}{2020.2021}\right)< \dfrac{1}{48}.\dfrac{1}{2}=\dfrac{1}{96}\)