Question

Tính giá trị của tổng sau : \(1.2.3+2.3.4+3.4.5+...+n\left(n+1\right)\left(n+2\right)\)

Answer 1

Đặt A là tên biểu thức

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

4A=1.2.3.4+2.3.4.4+...+n(n+1)(n+2).4

4A=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)

4A=[1.2.3.4+2.3.4.5+...+n(n+1)(n+2)(n+3)] - [0.1.2.3+1.2.3.4+...+(n-1)n(n+1)(n+2)]

4A=n(n+1)(n+2)(n+3)-0.1.2.3

A=\(\frac{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}{4}\)

Answer 2

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

\(\Rightarrow4A=1.2.3.4+2.3.4.4+3.4.5.4+...+4n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow4A=1.2.3.4+1.2.3.\left(5-1\right)+...+n\left(n+1\right)\left(n+2\right)\left(n+3-n+1\right)\)

\(\Rightarrow4A=1.2.3.4+2.3.4.5-1.2.3.4+...+n\left(n+1\right)\left(n+2\right)\left(n+3\right)-\left(n-1\right)n\left(n+1\right)\left(n\right)\)

\(\Rightarrow4A=n\left(n+1\right)\left(n+2\right)\left(n+3\right)\)

\(\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}{4}\)