Question

[Phạm Minh Tuấn]

Chứng minh : Với k thuộcN^* ta luôn có: k(k+1)(k+2)-(k-1)k(k+1)=3.k(k+1). Áp dụng tính tổng : S=1.2+2.3+3.4+...+n.(n+1)

Answer 1

[Quỳnh Giang Bùi]

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