A=1+(1+1).2+(2+1).3+...+(n−1+1).n⇔A=1+1.2+2+2.3+3+...+(n−1).n+n
⇔A=(1+2+3+...+n)+[1.2+2.3+3.4+...+(n−1).n]
Đặt B=1.2+2.3+3.4+...(n−1).n
⇔3B=1.2.3+2.3.(4−1)+3.4.(5−2)+....+(n−1)n[(n+1)−(n−2)]
⇔3B=1.2.3+2.3.4−1.2.3+3.4.5−2.3.4+....+(n−1)n(n+1)−(n−2)(n−1)n
⇔3B=(n−1)n(n+1)⇔B=(n−1)n(n+1)3
Do đó A=n(n+1)2+(n−1)n(n+1)3=n(n+1)(2n+1)6