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 Iris
By Iris
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)
Đáp án:
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)
⇒ 3(a1 + a2 + … + an) = n(n + 1)(n + 2)
⇒ 3[1.2+2.3+ … + n( n + 1 ) ] = n( n + 1 ) ( n + 2 )
⇒ A = $\frac{( n + 1 ) ( n + 2 )}{3}$
Bài 2 :
Ta có :
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)
⇒ B = $\frac{(n – 1)n(n + 1)(n + 2)}{4}$
$A=1.2+2.3+3.4+…+n(n+1)$
$⇒3A=1.2.3+2.3.4+3.4.3+…+n(n+1).3$
$⇒3A=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+…+n(n+1)[(n+2)-(n-1)$
$⇒3A=n(n+1)(n+2)$
$⇒A=$$\frac{n(n+1)(n+2)}{3}$
$B=1.2.3+2.3.4+…+(n-1)n(n+1)$
$⇒4B=1.2.3.4+2.3.4.4+…+(n-1)n(n+1).4$
$⇒4B=1.2.3.(4-0)=2.3.4.(5-1)+…+(n-1)n(n+1).[(n+2)-(n-2)$
$⇒4B=(n-1)n(n+1)(n+2)$
$⇒B=$$\frac{(n-1)n(n+1)(n+2)}{4}$