A=\(x = {n(n+1)(n+2){} \over 3}\)

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

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

Ta có các công thức:

1^2 + 2^2 + 3^2 +...+ n^2 = n(n+1)(2n+1)/6

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

Thay vào ta có:

S = n(n+1)(2n+1)/6 + n(n+1)/2

=n(n+1)/2[(2n+1)/3 + 1]

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

Bài 1 :

\(A=1\cdot2+2\cdot3+3\cdot4+...+n\cdot\left(n+1\right)\)

\(\Rightarrow3A=1\cdot2\cdot3+2\cdot3\cdot3+3\cdot4\cdot3+...+n\cdot\left(n+1\right)\cdot3\)

\(=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+n\cdot\left(n+1\right)\cdot\left[\left(n+2\right)-\left(n-1\right)\right]\)

\(=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+2\cdot3\cdot4-3\cdot4\cdot5+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)

\(=n\left(n+1\right)\left(n+2\right)\)

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

Bài 1.

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

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

3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + n.(n + 1).(n + 2 - n - 1)

3A = 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)

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

A = n.(n + 1).(n + 2) : 3

Bài 2. 

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 + 2.3.4.(5 - 1) + .... + (n - 1).n.(n + 1).(n + 2 - n - 2)

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

4B = (n - 1).n.(n + 1).(n + 2)

B = (n - 1).n.(n + 1).(n + 2) : 4

Xong rồi nhé anh !

Bài 2 :

\(B=1\cdot2\cdot3+2\cdot3\cdot4+...+\left(n-1\right)n\left(n+1\right)\)

\(\Rightarrow4B=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot4+...+\left(n-1\right)n\left(n+1\right)\cdot4\)

\(=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot\left(5-1\right)+...+\left(n-1\right)n\left(n+1\right)\left[\left(n+2\right)-\left(n-2\right)\right]\)

\(=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+...+\left(n-1\right)n\left(n+1\right)\left(n+2\right)-\left(n-2\right)\left(n-1\right)n\left(n+1\right)\)

\(=\left(n-1\right)n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow B=\frac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{4}\)

tick mk lên 70 điểm với các bạn

Bài 1:

\(A=1.2+2.3+3.4+...+n.\left(n+1\right)\)

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

\(=1.2\left(3-0\right)+2.3\left(4-1\right)+...+n.\left(n+1\right).\left[\left(n+2\right)-\left(n-1\right)\right]\)

=[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)] 

\(=n.\left(n+1\right).\left(n+2\right)\)

\(\Leftrightarrow A=\frac{\left[n.\left(n+1\right).\left(n+2\right)\right]}{3}\)

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) 

=>A=[n.(n+1).(n+2)] /3

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

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

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

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

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

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

bài 2 làm tương tự nhưng là 4B nha cậu

https://olm.vn/hoi-dap/tim-kiem?q=t%C3%ADnh+t%E1%BB%95ng+sau+:S+=+1.2.3+2.3.4+3.4.5+...+n.(n+1).(n+2)+&id=601088

Bài 1 Số số hạng của dãy là : (50-1):1+1=50(số hạng )

         S = (50+1) x 50 : 2 = 1275