Đặt A = 1.2 + 2.3 + 3.4 + .... + n(n + 1)

3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + n(n + 1).3

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

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

= n(n + 1)(n + 2)

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

A=1.2.3+2.3.3+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=1.2.3-1.2.0-2.3.4-2.3.1+......+N(N-1)+(N+2)-N(N-1)-N-1

3A=N(N-1)+(N+2)/3

Đặt S = 1.2 + 2.3 + 3.4 + .... + N( N + 1 )

3S = 1.2.3 + 2.3.3 + 3.4.3 + .... + N( N + 1 ).3

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

=> 1.2 .3 + 2.3.4 -1.2.3 + 3.4.5 - 2.3.4 + .... + N( N + 1 ) ( N - 1 )N( N + 1 )

=> N( N + 1 )( N + 2 )

=> S = \(\frac{N\left(N+1\right)\left(N+2\right)}{3}\)

Bài giải:

3S = 1.2.3 + 2.3.3 + 3.4.3 + ... + n(n +1)3

= 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) + ...+ n(n + 1)[(n + 2) - (n -1)]

= 1.2.3 + 2.3.4 - 2.3 + 3.4.5 - 2.3.4 + ... + n(n + 1)(n + 2) - n(n + 1)(n - 1)

= n(n + 1)(n + 2)

=> S N(N+1)(n+2)/3

3S = 1.2.3 + 2.3.3 + 3.4.3 + ... + n(n +1)3

= 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) + ...+ n(n + 1)[(n + 2) - (n -1)]

= 1.2.3 + 2.3.4 - 2.3 + 3.4.5 - 2.3.4 + ... + n(n + 1)(n + 2) - n(n + 1)(n - 1)

= n(n + 1)(n + 2)

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

Mk suýt nữa biết !

1.2 + 2.3 + 3.4 + ..... + n(n + 1)

\(=\frac{1.2.3+2.3.3+3.4.3+...+n\left(n+1\right).3}{3}\)

\(=\frac{1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+n\left(n+1\right)\left[\left(n+2\right)-\left(n-1\right)\right]}{3}\)

\(=\frac{1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+..+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)}{3}\)

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

Đặt A = 1.2 + 2.3 + 3.4 + ..... + n(n + 1)

3A = 1.2.3 + 2.3.3 + 3.4.3 + ..... + n(n + 1)3

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

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

= n(n + 1)(n + 2)

\(\Rightarrow A=\frac{N\left(N+1\right)\left(N+2\right)}{3}\)

3A=1.2.3+2.3.3+3.4.3+...+n(n+1).3

3A=1.2(3-0)+2.3(4-1)+3.4(5-3)+....+n(n+1)(n-2)-(n-1)

3A=1.2.3-1.2.0+2.3.4-2.3.3+.+n(n+1)+(n+2)-(n-1)+n(n-1)

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

3A=1.2.3+2.3.3+3.4.3+.....+N(N+1).3

3A=1.2(3-0)+2.3(4-1)+3.4(5-3)+........+n(n+1)(n-2)-(n-1)

3a=1.2.3-1.2.0+2.3.4-2.3.3+....+n(n+1)+(n+2)-(n-1)+n(n+1)

=>n(n-1)+(n+2)=n(n-1)+(n+2)/3

3A=1.2.3+2.3.3+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=1.2.3-1.2.0+2.3.4-2.3.1+......+N(N+1)+(N+2)-N-1+N(N-1)

=>N(N-1)+(N+2)=N(N-1)-(N+2)/3

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

S=1.2+2.3+3.4+.....+N(N+1)

3S=1.2(3-0)+2.3(4-1)+....+N.(N+1)+N(N+1)-(N+2)

3S=1.2.3+2.3.4+3.4.5+.+N(N+1)+(N+2)

3S=N(N+1)+(N+2)

S=N(N+1)+(N+2)/3

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 tk mình mk tk lại nha

Rảnh v~