Ta có:
A =1.2+2.3+3.4+.....+n(n+1)
=>3A=1.2.3+2.3.3+3.4.3+.....+n(n+1).3
3A=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+.....+n(n+1)[n+2-(n-1)]
3A=1.2.3-1.2.0+2.3.4-2.3.1+3.4.5-3.4.2+.....+n(n+1)(n+2)-n(n+1)(n-1)
3A=n(n+1)(n+2)-1.2.0
3A=n(n+1)(n+2)
A=\(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)