Question

Bài 1. Tính A = 1.2 + 2.3 + 3.4 + … + n.(n + 1)

Ai lm thì mik sẽ tik

Answer 1

Ta có: A = 1.2 + 2.3 + 3.4 + … + n.(n + 1)

   =>  3A = 1.2.(3-0) + 2.3.(4-1) + .... + n.(n+1).(n+2 - n+1)

   => 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .... + n.(n+1).(n+2)

  =>  3A = n.(n+1).(n+2)

  = > A = \(\frac{\text{n.(n+1).(n+2)}}{3}\)

Answer 2

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