Question

Chứng Minh \(D_n\) = 1.2+2.3+3.4+....+n(n+1) 

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

Answer 1

\(3D_n=1.2.3+2.3.3+3.4.3+...+n\left(n+1\right)3\)

\(=1.2\left(3-0\right)+2.3\left(4-1\right)+...+n\left(n+1\right)\left(n+2-n+1\right)\)

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

\(=n\left(n+1\right)\left(n+2\right)-0.1.2=n\left(n+1\right)\left(n+2\right)\)

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

Answer 2

D_n = 1.2 + 2.3 + 3.4 +...+ n(n + 1)

3D_n = 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) +...+ n(n + 1).[(n + 2) - (n - 1)]

3D_n = 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)

3D_n = [1.2.3 + 2.3.4 + 3.4.5 +...+ n(n + 1)(n + 2)] - [0.1.2 + 1.2.3 + 2.3.4 +...+ n(n - 1)(n + 1)]

3D_n = n(n + 1)(n + 2) - 0.1.2

3D_n = n(n + 1)(n + 2)

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