Question

tính 

1.2+2.3+3.4+4.5+...+n(n+1)

Answer 1

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

k cho mk nhé

Answer 2

đặt tổng trên là A

ta có:

3A=1.2.3+2.3.3+...+n.(n+1).3

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