Ta có : S = 1.2 + 2.3 + 3.4 + ..... + 99.100

=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .... + 99.100.101

=> 3S = 99.100.101

=> S = \(\frac{99.100.101}{3}=333300\)

ta xét

\(S\left(n\right)=1.2+2.3+..+n\left(n-1\right)\)

\(\Rightarrow3S\left(n\right)=1.2.3+2.3.3+..+3.n.\left(n-1\right)\)

\(\Leftrightarrow3S\left(n\right)=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+..+n\left(n-1\right)\left(n+1-\left(n-2\right)\right)\)

\(\Leftrightarrow3S\left(n\right)=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+..+n\left(n-1\right)\left(n+1\right)-n\left(n-1\right)\left(n-2\right)\)

\(\Leftrightarrow3S\left(n\right)=n\left(n-1\right)\left(n+1\right)\Rightarrow S\left(n\right)=\frac{n\left(n-1\right)\left(n+1\right)}{3}\)

Áp dụng ta có \(S\left(100\right)=\frac{99.100.101}{3}=333300\)

Em tham khảo:

Câu hỏi của nguyễn huy bảo - Toán lớp 7 - Học trực tuyến OLM

Ai mà bt đc

3s=1.2(3-0)+2.3(4-1)+...+n.(n+1).(n+2)-(n-1)

3a=1.2.3+2.3.4+3.4.5+....+(n-1).n(n+1)+n(n+1)(n+2)-

0.1..2-1.2.3-3.4.5-....-(n-1)n(n+1)

3a=n.(n+1)(n+2)

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

1.2+2.3+3.4.....+n.(n+1)=A 
ta có 
3.A=1.2.(3-0)+2.3.(4-1)+3.4.(5 -2)...+ n.(n+1) . ((n+2) - (n-1)) 
3.A=1.2.3+2.3.4+3.4.5+...+ (n-1) . n. (n+1)+ n. (n+1). (n+2) - 
0.1.2 -1.2.3 -2.3.4 -3.4.5 -...(n-1)n(n+1) 
3A=n.(n+1).(n+2) 
A=n.(n+1).(n+2)\3 

ai mình nt mik lại

k zồi , k tui

3S₂=1*2*(3-0)+2*3*(4-1)+...+ n*(n+1)*[(n+2)-(n-1)]

3S₂=1*2*3+2*3*4+...+n*(n+1)*(n+2)-0*1*2-1*2*3-...-(n-1)*n*(n+1)

3S₂=n*(n+1)*(n+2)

\(S_2=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

1.2+2.3+3.4.....+n.(n+1)=A 
ta có 
3.A=1.2.(3-0)+2.3.(4-1)+3.4.(5 -2)...+ n.(n+1) . ((n+2) - (n-1)) 
3.A=1.2.3+2.3.4+3.4.5+...+ (n-1) . n. (n+1)+ n. (n+1). (n+2) - 
0.1.2 -1.2.3 -2.3.4 -3.4.5 -...(n-1)n(n+1) 
3A=n.(n+1).(n+2) 
A=n.(n+1).(n+2)\3