Bài 1. Tính A = 1.2 + 2.3 + 3.4 + … + n.(n + 1) Bài 2. Tính B = 1.2.3 + 2.3.4 + … + (n – 1)n(n + 1)

Question

Bài 1. Tính A = 1.2 + 2.3 + 3.4 + … + n.(n + 1)
Bài 2. Tính B = 1.2.3 + 2.3.4 + … + (n – 1)n(n + 1)

in progress 0
Skylar 4 tuần 2021-09-16T18:31:36+00:00 2 Answers 7 views 0

Answers ( )

    0
    2021-09-16T18:32:54+00:00

    Bài 1 :

    Gọi a1 = 1.2 → 3a1 = 1.2.3 → 3a1 = 1.2.3 – 0.1.2
    a2 = 2.3 → 3a2 = 2.3.3 → 3a2 = 2.3.4 – 1.2.3
    a3 = 3.4 → 3a3 = 3.3.4 → 3a3 = 3.4.5 – 2.3.4
    …………………..
    an-1 = (n – 1)n → 3an-1 =3(n – 1)n → 3an-1 = (n – 1)n(n + 1) – (n – 2)(n – 1)n
    an = n(n + 1) → 3an = 3n(n + 1) → 3an = n(n + 1)(n + 2) – (n – 1)n(n + 1)

    Bài 2 :

    4B = 1.2.3.4 + 2.3.4.4 + … + (n – 1)n(n + 1).4

    = 1.2.3.4 – 0.1.2.3 + 2.3.4.5 – 1.2.3.4 + … + (n – 1)n(n + 1)(n + 2) – [(n – 2)(n – 1)n(n + 1)]

    = (n – 1)n(n + 1)(n + 2) – 0.1.2.3 = (n – 1)n(n + 1)(n + 2)

     

    0
    2021-09-16T18:33:35+00:00

    Đáp án:

     Bài 1:

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

    $⇔ 3A = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) +……+ n.(n+1).(n+2-n-1)$

    $⇔ 3A = 1.2.3 + 2.3.4 – 1.2.3 +….+ n.(n+1).(n+2) – (n-1).n.(n+1)$

    $⇔ 3A = n.(n+1).(n+2)$

    $⇔ A = \frac{n.(n+1).(n+2)}{3}$ 

    Bài 2:

    $B = 1.2.3 + 2.3.4 + … + (n – 1)n(n + 1)$

    $⇔ 4B = 1.2.3.4 + 2.3.4.(5-1) + … + (n – 1)n(n + 1)(n+2-n-2)$

    $⇔ 4B = 1.2.3.4 + 2.3.4.5 – 1.2.3.4 +……+ (n – 1)n(n + 1)(n+2) – (n-2)(n – 1)n(n + 1)$

    $⇔ 4B = (n – 1)n(n + 1)(n+2)$

    $⇔ B = \frac{(n – 1)n(n + 1)(n+2) }{4}$ 

     

Leave an answer

Browse

35:5x4+1-9:3 = ? ( )