Question

[zeno-sama]

tính B=1.2.3+2.3.4+...+(n-1)n(n+1) giusp mik nha.

Answer 1

[Nguyễn Thanh Hằng]

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

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

\(\Leftrightarrow4B=\left(4-0\right).1.2.3+\left(5-1\right).2.3.4+.........+\left[\left(n+2\right)-\left(n-2\right)\right]\left(n-1\right)n\left(n+1\right)\)

\(\Leftrightarrow4B=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+.......+\left(n-1\right)n\left(n+1\right)\left(n+2\right)\left(n+3\right)-\left(n-2\right)\left(n-1\right)n\left(n+1\right)\)

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

\(\Leftrightarrow B=\dfrac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{4}\)

Answer 2

[Nguyễn Trần Trúc Ly]

Ta có:

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

=> 4B=1.2.3.4+2.3.4.(5-1)+...+(n-1)n(n+1)((n+2)-(n-2))

=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+(n-1)n(n+1)(n+2)-(n-2)(n-1)n(n+1)

=(n-1)n(n+1)(n+2)

=> B=\(\dfrac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{4}\)

Vậy B=\(\dfrac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{4}\)