Ta có : k(k+1)(k+2)-(k-1)(k+1)k

         =k(k+1).[(k+2)-(k-1)]

         =3k(k+1)

áp dụng  3(1+2)=1.2.3-0.1.2

             =>3(2.3)=2.3.4-1.2.3

             =>3(3.4)=3.4.5-2.3.4

            .....................................

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

Cộng lại ta có   3.S=n(n+1)(n+2)=>S=n(n+1)(n+2)/3

CHÚC BẠN HỌC TỐT NHA !!!

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

S=1.2+2.3+3.4+...+n(n+1)

=>3S=1.2.3+2.3.3+3.4.3+...+n(n+1)3

=1.2.3+2.3.(4-1)+3.4(5-2)+...+n.(n+1)[(n+2)-(n-1)]

=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+n(n+1)(n+2)-(n-1)n(n+1)

=n(n+1)(n+2)

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

m tưởng tao thik đăng à..............................................

Ta có: 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)[(k-k)+(2+1)]

        =k(k+1)3

        =3k(k+1)

 Vậy k(k+1)(k+2)-(k-1)k(k+1)=3k(k+1)

Áp dụng:

  S=1.2+2.3+3.4+...+n(n+1)

3S=3.1.2+3.2.3+3.3.4+...+3.n(n+1)

3S=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+...+n(n+1)(n+2)-(n-1)n(n+1)

3S=(1.2.3-1.2.3)+(2.3.4-2.3.4)+(3.4.5-3.4.5)+...+[(n-1)n(n+1)-(n-1)n(n+1)]+n(n+1)(n+2)-0

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

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

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

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

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

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

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

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

Ta có:

k(k+1)(k+2)-(k-1)k(k+1)=k.(k+1).[(k+2)-(k-1)]

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

                                 =3k.(k+1)

Phần 2 đề sai phải là tính S=1.2.3+2.3.4+...+n.(n+1).(n+2)