\(A=1\cdot2+2\cdot3+...+n\left(n+1\right)\)
=>\(3\cdot A=1\cdot2\cdot3+2\cdot3\cdot3+...+3n\left(n+1\right)\)
=>\(3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+n\left(n+1\right)\left[\left(n+2\right)-\left(n-1\right)\right]\)
=>\(3A=1\cdot2\cdot3-1\cdot2\cdot3+2\cdot3\cdot4-2\cdot3\cdot4+...+n\left(n+1\right)\left(n-1\right)-n\left(n+1\right)\left(n-1\right)+n\left(n+1\right)\left(n+2\right)\)
=>\(3A=n\left(n+1\right)\left(n+2\right)\)
=>\(A=\dfrac{n\left(n+1\right)\left(n+2\right)}{3}\)
Ta có : A = 1.2 + 2.3 + 3.4 + … + n.(n + 1)
⇒3A = 1.2.(3-0)+2.3.(4-1)+3.4.(5-2).....n.(n+1).[(n+2)-(n-1)]
⇒3A = 1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+....+n.(n+1)(n+2)-(n-1)n(n+1)
⇒3A = (1.2.3-1.2.3)+(2.3.4-2.3.4)+....+[(n-1).n.(n+1)-(n-1)n(n+1)]+n.(n+1)(n+2)
⇒3A = n.(n+1)(n+2)
⇒A = n.(n+1)(n+2) / 3
