Question

tính tổng \(S=1.2+2.3+...+\left(n-2\right)\left(n-1\right)+\left(n-1\right)n\)

Answer 1

Ta có : \(3S=1.2.\left(3-0\right)+2.3.\left(4-1\right)+...+\left(n-2\right)\left(n-1\right)\left[n-\left(n-3\right)\right]+\left(n-1\right)n.\left[\left(n+1\right)-\left(n-2\right)\right]\)

\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+\left(n-2\right)\left(n-1\right)n-\left(n-3\right)\left(n-2\right)\left(n-1\right)+\left(n-1\right)n\left(n+1\right)-\left(n-2\right)\left(n-1\right)n\)

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

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

Vậy : \(S=\frac{\left(n-1\right)n\left(n+1\right)}{3}\)