We have:

k(k+1)(k+2)-(k-1)k(k+1)=k(k+1)(k+2-k+1)=3k(k+1)

=>k(k+1)=\(\dfrac{k\left(k+1\right)\left(k+2\right)}{3}\)-\(\dfrac{\left(k-1\right)k\left(k+1\right)}{3}\)

k(k+1)(k+2)−(k−1)k(k+1)k(k+1)(k+2)−(k−1)k(k+1)

=k(k+1)(k+2−k+1)=k(k+1)(k+2−k+1)

=k(k+1)⋅3=k(k+1)⋅3

=3k(k+1)

Dao Trong Luan is correct

\(k\left(k+1\right)\left(k+2\right)-\left(k-1\right)k\left(k+1\right)\)

\(=k\left(k+1\right)\left(k+2-k+1\right)\)

\(=k\left(k+1\right)\cdot3\)

\(=3k\left(k+1\right)\)

Are you OK?