Question

Tính tổng A= 1.2 + 2 .3 + 3 .4 +...+(n - 1 )n .( 1)

Answer 1

3A=1.2.3+2.3.(4-1)+3.4.(5-2)+...+(n-1)n[(n+1)-(n-2)]
3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+(n-1)n(n+1)-(n-2)(n-1)n
3A=(n-1)n(n+1)
A=(n-1)n(n+1)/3

Answer 2

Ta có : 

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

\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+...+\left(n-1\right).n.3\)

\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+\left(n-1\right).n.\left[\left(n+1\right)-\left(n-2\right)\right]\)

\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+\left(n-1\right)n.\left(n+1\right)-\left(n-2\right).\left(n-1\right).n\)

\(\Rightarrow3A=\left(n-1\right).n.\left(n+1\right)\)

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

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

P/s : Mik ko chắc 

~ Ủng hộ nhé 

Answer 3

ta có: \(A=1.2+2.3+3.4+...+\left(n-1\right).n\)

\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+...+\left(n-1\right).n.3\)

\(3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+\left(n-1\right).n.\left[\left(n+1\right)-[\left(n-1\right)-1]\right]\)

\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.34+...+\left(n-1\right).n.\left(n+1\right)-\left[\left(n-1\right)-1\right].\left(n-1\right).n\)

\(3A=\left(n-1\right).n.\left(n+1\right)\)

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