Nguyễn Tuấn Minh
Giới thiệu về bản thân
3A=1.2.3+2.3.3+3.4.3+...+n.(n+1).3 3A=1.2.3+2.3.(4-1)+3.4.(5-2)+...+(n-1).n.[(n+1)-(n-2)]+n.(n+1).[(n+2)-(n-1)] 3A=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+...-(n-2).(n-1).n+(n-1).n.(n+1)- (n-1).n.(n+1) + n.(n+1).(n+2) 3A=n.(n+1).(n+2) A=\(\frac{n.\left(n+1\right).\left(n+2\right)}{3}\)