Question

a) Tính tổng: 1+2+3+...+n, 1+3+5+...+(2n-1)

b) Tính tổng: 1.2+2.3+3.4+...+n(n+1)

                   1.2.3+2.3.4+3.4.5+...+n(n+1)(n+2)

Với n là số tự nhiên khác không

Answer 1

a)  A =(2n-1+1).(2n-1)/2=2n.(2n-1)/2=n(2n-1)

b)  B= 1.2+2.3+3.4+...+n(n+1)

3B=1.2.3+2.3.(4-1)+3.4.(5-2)+...+n(n+1)[(n+2)-(n-1)]

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

3B=n(n+1)(n+2)

B=n(n+1)(n+2)/3

 

4C=1.2.3.4+2.3.4.(5-1)+3.4.5(6-2)+...+n(n+1)(n+2).[(n+3)-(n-1)]

4C=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+...+n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)

4C=n(n+1)(n+2)(n+3)

C=n(n+1)(n+2)(n+3)/4

Answer 2

Câu b1 nếu mà là (n-1) thì sao

Answer 3

B,C lay o dau vay

Answer 4

3B=1.2.3+2.3.(4-1)+3.4.(5-2)+...+n(n+1)[(n+2)-(n-1)] 3B=1.2.3-1.2.3+2.3.4-2.3.4+...+n(n+1)(n+2)-(n-1)n(n+1) 3B=n(n+1)(n+2) B=n(n+1)(n+2)/3 4C=1.2.3.4+2.3.4.(5-1)+3.4.5(6-2)+...+n(n+1)(n+2).[(n+3)-(n-1)] 4C=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+...+n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2) 4C=n(n+1)(n+2)(n+3) C=n(n+1)(n+2)(n+3)/4