Tính a.1*2+2*3+3*4+….+n*(n+1) b.1*2*3+2*3*4+3*4*5+…..n*(n+1)*(n+2) 25/09/2021 Bởi Josie Tính a.1*2+2*3+3*4+….+n*(n+1) b.1*2*3+2*3*4+3*4*5+…..n*(n+1)*(n+2)
a) 3A=1.2.3 + 2.3.3 + 3.4.3 +… + n.(n+1).3 =1.2.(3-0) + 2.3.(4-1) + … + n.(n+1).[(n+2)-(n-1)] =[1.2.3+ 2.3.4 + …+ (n-1).n.(n+1)+ n.(n+1)(n+2)] – [0.1.2+ 1.2.3 +…+(n-1).n.(n+1)] =n.(n+1).(n+2) =>S=[n.(n+1).(n+2)] /3 b) Nhân 4 vào hai vế ta được: 4A = 4.[1.2.3 + 2.3.4 + 3.4.5 + … + (n – 1).n.(n + 1)] 4A = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + … + (n – 1).n.(n + 1).4 4A = 1.2.3.4 + 2.3.4.(5 – 1) + 3.4.5.(6 – 2) + … + (n – 1).n.(n + 1).[(n + 2) – (n – 2)] 4A = 1.2.3.4 + 2.3.4.5 – 1.2.3.4 + 3.4.5.6 – 2.3.4.5 + … + (n – 1).n(n + 1).(n + 2) – (n – 2).(n – 1).n.(n + 1) 4A = (n – 1).n(n + 1).(n + 2) A = (n – 1).n(n + 1).(n + 2) : 4. 3A=1.2.3 + 2.3.3 + 3.4.3 +… + n.(n+1).3 =1.2.(3-0) + 2.3.(4-1) + … + n.(n+1).[(n+2)-(n-1)] =[1.2.3+ 2.3.4 + …+ (n-1).n.(n+1)+ n.(n+1)(n+2)] – [0.1.2+ 1.2.3 +…+(n-1).n.(n+1)] =n.(n+1).(n+2) =>S=[n.(n+1).(n+2)] /3 Bình luận
a) 3A=1.2.3 + 2.3.3 + 3.4.3 +… + n.(n+1).3
=1.2.(3-0) + 2.3.(4-1) + … + n.(n+1).[(n+2)-(n-1)]
=[1.2.3+ 2.3.4 + …+ (n-1).n.(n+1)+ n.(n+1)(n+2)] – [0.1.2+ 1.2.3 +…+(n-1).n.(n+1)]
=n.(n+1).(n+2)
=>S=[n.(n+1).(n+2)] /3
b)
Nhân 4 vào hai vế ta được:
4A = 4.[1.2.3 + 2.3.4 + 3.4.5 + … + (n – 1).n.(n + 1)]
4A = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + … + (n – 1).n.(n + 1).4
4A = 1.2.3.4 + 2.3.4.(5 – 1) + 3.4.5.(6 – 2) + … + (n – 1).n.(n + 1).[(n + 2) – (n – 2)]
4A = 1.2.3.4 + 2.3.4.5 – 1.2.3.4 + 3.4.5.6 – 2.3.4.5 + … + (n – 1).n(n + 1).(n + 2) – (n – 2).(n – 1).n.(n + 1)
4A = (n – 1).n(n + 1).(n + 2)
A = (n – 1).n(n + 1).(n + 2) : 4.
3A=1.2.3 + 2.3.3 + 3.4.3 +… + n.(n+1).3
=1.2.(3-0) + 2.3.(4-1) + … + n.(n+1).[(n+2)-(n-1)]
=[1.2.3+ 2.3.4 + …+ (n-1).n.(n+1)+ n.(n+1)(n+2)] – [0.1.2+ 1.2.3 +…+(n-1).n.(n+1)]
=n.(n+1).(n+2)
=>S=[n.(n+1).(n+2)] /3