Tính: A=1.2+2.3+3.4+...+n(n+1)
=> 3A= 1.2.3+ 2.3.4+ .......+ 3n.( n+1)
=> 3A= 1.2.3 + 2.3.4- 1.2.3 + 3.4.5- 2.3.4 +......+ n(n+ 1) . ( n+ 2)- n. (n-1) .( n+1)
=> 3A= n( n+1) . (n+2)
=> A= \(\dfrac{n\left(n+1\right).\left(n+2\right)}{3}\)
Vậy A = \(\dfrac{n\left(n+1\right).\left(n+2\right)}{3}\) \(⋮\)3
A = 1.2 + 2.3 + 3.4 +...+ n.(n+1)
3A=1.2.3 + 2.3.3 + 3.4.3 +... + n.(n+1).3
=1.2.(3-0) + 2.3.(4-1) + ... + n.(n+1).[(n+2)-(n-1)]
=[1.2.3+ 2.3.4 + ...+ (n-1).n.(n+1)+ n.(n+1)(n+2)] - [0.1.2+ 1.2.3 +...+(n-1).n.(n+1)]
=n.(n+1).(n+2)
=>A=[n.(n+1).(n+2)] /3
