Question

[Tập-chơi-flo]

Hãy tính tổng 

C = 1.2.3+ 2.3.4 + 3.4.5 +...+n(n+1) ( n+2)

Answer 1

[Incursion_03]

C = 1.2.3+ 2.3.4 + 3.4.5 +...+n(n+1) ( n+2)

\(\Rightarrow4C=1.2.3\left(4-0\right)+2.3.4.\left(5-1\right)+...+n\left(n+1\right)\left(n+2\right)\left[\left(n+3\right)-\left(n-1\right)\right]\)

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

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

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

Answer 2

[Tập-chơi-flo]

Thanks