Ta có:

\(k\left(k+1\right)\left(k+2\right)-\left(k-1\right)k\left(k+1\right)\\ =k\left(k+1\right)\left[\left(k-2\right)-\left(k-1\right)\right]\\ =k\left(k+1\right)\left[k-2-k+1\right]\\ =k\left(k+1\right)\left\{\left[k+\left(-k\right)\right]+\left(2+1\right)\right\}\\ =k\left(k+1\right).3\\ =3.k\left(k+1\right)\)

Vậy \(k\left(k+1\right)\left(k+2\right)-\left(k-1\right)k\left(k+1\right)\\ =3.k.\left(k+1\right)\)

Ta có:

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

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

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

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

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

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

\(\Rightarrow VT=VP\)

Vậy với \(k\in N\)* thì ta luôn có:

\(k\left(k+1\right)\left(k+2\right)-\left(k-1\right)k\left(k+1\right)=3k\left(k+1\right)\) (Đpcm)

hinh nhu de sai

\(k\left(k+1\right)\left(k+2\right)-\left(k-1\right)k\left(k+1\right)=k\left(k+1\right)\left[\left(k+2\right)-\left(k-1\right)\right]=3k\left(k+1\right)\)

Công thức tinh tổng là : \(S=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

\(k\left(k+1\right)\left(k+2\right)-\left(k-1\right)k\left(k+1\right)=k\left(k+1\right)\left(k+2-k+1\right)=3k\left(k+1\right)\left(ĐPCM\right)\)

\(S=1.2+2.3+3.4+...+n\left(n+1\right)\)

3\(S=3\left[1.2+2.3+3.4+...+n\left(n+1\right)\right]\)

\(3S=1.2.3-0.1.2+2.3.4-1.2.3+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)

3S=n(n+1)(n+2)

\(S=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

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

=(k+1)(k²+2k)-(k²-k)(k+1)

=(k+1)[(k²+2k)-(k²-k)]

=(k+1)[k²+2k-k²+k]

=(k+1)[(k²-k²)+(2k+k)]

=(k+1)3k (Đpcm)

4S=1.2.3.4+2.3.4.4+3.4.5.4+...+k(k+1)(k+2).4=

=1.2.3.4+2.3.4(5-1)+3.4.5.(6-2)+...+k(k+1)(k+2)[(k+3)-(k-1)]=

=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-...-(k-1)k(k+1)(k+2)+k(k+1)(k+2)(k+3)=

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

=(k²+3k)(k²+3k+2)=(k²+3k)²+2(k²+3k)

=> 4S+1=(k²+3k)²+2(k²+3k)+1=[(k²+3k)+1]²