Question

Tính A = 1.2 + 2.3 + 3.4 + … + n.(n + 1)

Answer 1

3A=1.2.3+2.3.(4-1)+.............+n.(n+1).[(n+2)-(n-1)]

3A=1.2.3+2.3.4-1.2.3+............+n.(n+1).(n+2)-(n-1).n.(n+1)

3A=n.(n+1).(n+2)

A=\(\frac{n.\left(n+1\right).\left(n+2\right)}{3}\)