Question

Chứng minh : Với k E N* ta luôn có :

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

Answer 1

Ta có : \(k\left(k+1\right)\left(k+2\right)-\left(k-1\right)k\left(k+1\right)\)

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

\(=k^3+2k^2+k^2+2k-k^3+k\)

\(=3k^2+3k\)

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

\(\Rightarrowđpcm\)

Answer 2

k(k+1)(k+2) -(k-1)k(k+1)

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

= 3k(k+1)  đpcm