Giải thích các bước giải:
Ta có:
$A=\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{(3n-1)(3n+2)}$
$\to 3A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{(3n-1)(3n+2)}$
$\to 3A=\dfrac{5-2}{2.5}+\dfrac{8-5}{5.8}+...+\dfrac{3n+2-(3n-1)}{(3n-1)(3n+2)}$
$\to 3A=\dfrac12-\dfrac15+\dfrac15-\dfrac18+...+\dfrac1{3n-1}-\dfrac1{3n+2}$
$\to 3A=\dfrac12-\dfrac1{3n+2}$
$\to A=\dfrac16-\dfrac1{3(3n+2)}$
$\to A=\dfrac{n}{2(3n+2)}$
Vì $n\in N , n>1\to n<2(3n+2)\to \dfrac{n}{2(3n+2)}<1\to A<1$
Mà $A>0$
$\to 0<A<1$
$\to 2020<2020+A<2021$
Lại có $M=2020+A$
$\to 2020<M<2021$
$\to M$ không là số nguyên
