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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 7: Bài 1. Tính A = 1.2 + 2.3 + 3.4 + … + n.(n + 1) Nhanh lên nha sắp nộp r

Toán Lớp 7: Bài 1. Tính A = 1.2 + 2.3 + 3.4 + … + n.(n + 1)
Nhanh lên nha sắp nộp r

Comments ( 2 )

  1.         A = 1.2 + 2.3 + 3.4 + … + n.(n + 1)
    => 3A = 1.2.3 + 2.3.3 + … + n(n + 1).3
               = 1.2.(3 – 0) + 2.3.(3 – 1) + … + n(n + 1)[(n – 2) – (n – 1)]
                  = 1.2.3 – 1.2.0 + 2.3.3 – 1.2.3 + … + n(n + 1)(n + 2) – (n – 1)n(n + 1)
                    = n(n + 1)(n + 2)
    Vậy A = $\frac{n(n + 1)(n + 2)}{3}$ 

  2. Giải:
    A = 1.2 + 2.3 + 3.4 + … + n(n + 1)
    -> 3A = 1.2.3 + 2.3.3 + 3.4.3 + … + 3n(n + 1)
    -> 3A = 1.2.3 + 2.3.(4 – 1) + 3.4(5 – 2) + … + [(n + 1) – (n – 2)]n(n -1)
    -> 3A = 2.3 + 2.3.4 – 2.3 + 3.4.5 – 2.3.4 + … + (n + 1)n(n – 1) – (n – 2)n(n – 1)
    -> 3A = (n – 1)n(n + 1)
    -> A = ((n – 1)n(n + 1))/3
     

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222-9+11+12:2*14+14 = ? ( )