Question

Tính B = 1.2.3 + 2.3.4 + ... + (n - 1)n(n + 1)

giúp mình giải bài này nha.thank(bn nào tl nhanh nhất mk sẽ tik cho nha)

Answer 1

\(B=1.2.3+2.3.4+...+\left(n-1\right).n.\left(n+1\right)\)

\(4B=1.2.3.4+2.3.4.\left(5-1\right)+...+\left(n-1\right).n.\left(n+1\right)\left[\left(n+2\right)-\left(n-2\right)\right]\)

\(4B=1.2.3.4+2.3.4.5-1.2.3.4+...+\left(n-1\right).n.\left(n+1\right)\left(n+2\right)-\left(n-2\right)\left(n-1\right).n.\left(n+1\right)\)

\(4B=\left(n-1\right).n.\left(n+1\right)\left(n+2\right)\)

\(B=\frac{\left(n-1\right).n.\left(n+1\right)\left(n+2\right)}{4}\)

Tham khảo nhé~

Answer 2

Ta có: \(B=1.2.3+2.3.4+...+\left(n-1\right).n.\left(n+1\right)\)

\(\Leftrightarrow4B=4.\left[1.2.3+2.3.4+...+\left(n-1\right).n.\left(n+1\right)\right]\)

\(\Leftrightarrow4B=1.2.3.4+2.3.4.4+...+\left(n-1\right).n.\left(n+1\right).4\)

\(\Leftrightarrow4B=1.2.3.4+2.3.4\left(5-1\right)+...+\left(n-1\right)n.\left(n+1\right).\left[\left(n+2\right)-\left(n-2\right)\right]\)

\(\Leftrightarrow4B=1.2.3.4+2.3.4.5-1.2.3.4+...+\left(n-1\right).n.\left(n+1\right).\left(n+2\right)-\left(n-2\right).\)\(\left(n-1\right).n.\left(n+1\right)\)

\(\Leftrightarrow4B=\left(n-1\right).n.\left(n+1\right).\left(n+2\right)\)

\(\Leftrightarrow B=\left(n-1\right).n.\left(n+1\right).\left(n+2\right)\div4\)

Vậy \(B=\left(n-1\right).n.\left(n+1\right).\left(n+2\right)\div4\)