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)
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)
By Skylar
By Skylar
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)
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)
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:
$B = 1.2.3 + 2.3.4 + … + (n – 1)n(n + 1)$
$⇔ 4B = 1.2.3.4 + 2.3.4.(5-1) + … + (n – 1)n(n + 1)(n+2-n-2)$
$⇔ 4B = 1.2.3.4 + 2.3.4.5 – 1.2.3.4 +……+ (n – 1)n(n + 1)(n+2) – (n-2)(n – 1)n(n + 1)$
$⇔ 4B = (n – 1)n(n + 1)(n+2)$
$⇔ B = \frac{(n – 1)n(n + 1)(n+2) }{4}$