Question

(2n+3)^2 - (2n-1)^2 chia hết cho 8 với n thuộc Z

Answer 1

`(2n+3)^2-(2n-1)^2`

`=(2n+3+2n-1)(2n+3-2n+1)`

`=(4n+2).4`

`=8.(2n+1) vdots 9 forall n \in ZZ`

Answer 2

Ta có: \(\left(2n+3\right)^2-\left(2n-1\right)^2\)

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

\(=\left(4n-2\right)\cdot4⋮8\)