Question

tính tổng B=1.2.3+2.3.4+3.4.5+......+n(n+1)(n+2)

Answer 1

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

  ={1.2.3.(4-0)+2.3.4(5-1)+3.4.5.(6-2)+...+n(n+1)(n+2)[(n+3)-(n-1)]} : 4

  = [1.2.3.4+2.3.4.5+3.4.5.6+...+n(n+1)(n+2)(n+3) - 1.2.3.4 - 2.3.4.5 - 3.4.5.6 - ... - n(n+1)(n+2)(n-1)] : 4

  =\(\frac{\text{ n(n+1)(n+2)(n+3) }}{4}\)

 

Answer 2

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