Question

Tính tổng:

\(1.2+2.3+3.4+...+n\left(n+1\right)\)

\(1.2.3+2.3.4+3.4.5+...+n\left(n+1\right)\left(n+2\right)\)

Answer 1

Ribi Nkok Ngok''>

Answer 2

Gọi A=

4A=1.2.3+2.3.4+3.4.5+...+n(n+1)(n+2)

=> 4A=1.2.3(4-0)+2.3.4(5-1)+...+n(n+1)(n+2)[(n+3)-(n-1)]

=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+...+n(n+1)(n+2)(n+3)-(n-1).n(n+1)(n+2)

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

=>

Answer 3

YNguyễn Thị Diễm QuỳnhHoàng Tử HàAkai Haruma

Answer 4

\(3A=1\cdot2\cdot\left(3-0\right)+2\cdot3\cdot\left(4-1\right)+...+n\left(n+1\right)\left[\left(n+2\right)-\left(n-1\right)\right]\)

\(3A=1\cdot2\cdot3-0\cdot1\cdot2+2\cdot3\cdot4-1\cdot2\cdot3+...+\)\(n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)

\(3A=n\left(n+1\right)\left(n+2\right)-0\cdot1\cdot2\)

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

câu còn lại nhân 4 rồi lm tương tự