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

=>\(4B=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot4+...+\left(n-1\right)\cdot n\left(n+1\right)\cdot4\)

=>\(4B=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\left(5-1\right)+...+\left(n-1\right)\cdot n\left(n+1\right)\left[\left(n+2\right)-\left(n-2\right)\right]\)

=>\(4B=1\cdot2\cdot3\cdot4-1\cdot2\cdot3\cdot4+...+\left(n-2\right)\left(n-1\right)\cdot n\cdot\left(n+1\right)-\left(n-2\right)\cdot\left(n-1\right)\cdot n\cdot\left(n+1\right)+\left(n-1\right)\cdot n\left(n+1\right)\left(n+2\right)\)

=>\(4B=\left(n-1\right)\cdot n\cdot\left(n+1\right)\left(n+2\right)\)

=>\(B=\dfrac{\left(n-1\right)\cdot n\left(n+1\right)\left(n+2\right)}{4}\)

\(C=1\cdot4+2\cdot5+3\cdot6+...+n\left(n+3\right)\)

\(=1\cdot\left(1+3\right)+2\left(2+3\right)+...+n\left(n+3\right)\)

\(=\left(1^2+2^2+...+n^2\right)+3\left(1+2+...+n\right)\)

\(=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}+3\cdot\dfrac{n\left(n+1\right)}{2}\)

\(=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}+\dfrac{3n\left(n+1\right)}{2}\)

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

\(=\dfrac{n\left(n+1\right)}{2}\cdot\dfrac{2n+1+9}{3}\)

\(=\dfrac{n\left(n+1\right)\left(n+5\right)}{3}\)

\(D=1^2+2^2+...+n^2\)

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

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

Đặt \(A=1+2+3+...+n;E=1\cdot2+2\cdot3+...+\left(n-1\right)\cdot n\)

\(E=1\cdot2+2\cdot3+...+\left(n-1\right)\cdot n\)

=>\(3E=1\cdot2\cdot3+2\cdot3\cdot3+...+\left(n-1\right)\cdot n\cdot3\)

=>\(3E=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+\left(n-1\right)\cdot n\left[\left(n+1\right)-\left(n-2\right)\right]\)

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

=>\(3E=\left(n-1\right)\cdot n\left(n+1\right)=n^3-n\)

=>\(E=\dfrac{n^3-n}{3}\)

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

Số số hạng là n-1+1=n(số)

Tổng của dãy số là: \(A=\dfrac{n\left(n+1\right)}{2}\)

=>\(D=\dfrac{n^3-n}{3}+\dfrac{n\left(n+1\right)}{2}\)

\(=\dfrac{2n^3-2n+3n^2+3n}{6}\)

=>\(D=\dfrac{2n^3+3n^2+n}{6}\)

Ta thấy:

1.4 = 1.(1 + 3) = 1.(1 + 1 + 2) = 1.(1 + 1)+ 2.1

2.5 = 2.(2 + 3) = 2.(2 + 1 + 2) = 2.(2 + 1)+ 2.2

3.6 = 3.(3 + 3) = 3.(3 + 1 + 2) = 3.(3 + 1)+ 2.3

4.7 = 4.(4 + 3) = 4.(4 + 1 + 2) = 4.(4 + 1)+ 2.4

. . . . . . . . . . .

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

Vậy C = 1.2 + 2.1 + 2.3 + 2.2 + 3.4 + 2.3 + . . . + n(n + 1) + 2n

= 1.2 + 2 +2.3 + 4 + 3.4 + 6 + . . . + n(n + 1) + 2n

= [1.2 +2.3 +3.4 + . . . + n(n + 1)] + (2 + 4 + 6 + . . . + 2n)

Mà 1.2 + 2.3 + 3.4 + … + n.(n + 1) =\(\frac{n.\left(n+1\right).\left(n+2\right)}{3}\)

Và 2 + 4 + 6 + . . . + 2n =\(\frac{\left(2n+2\right).n}{2}\)

=> C=\(\frac{n.\left(n+1\right).\left(n+2\right)}{3}+\frac{\left(2n+2\right).n}{2}-\frac{n.\left(n+1\right).\left(n+5\right)}{3}\)

hok tốt

Ta có : 

\(C=1.4+2.5+3.6+...+n\left(n+3\right)\)

\(\Rightarrow C=1\left(2+2\right)+2\left(3+2\right)+3\left(4+2\right)+...+n\left(n+1+2\right)\)

\(\Rightarrow C=1.2+1.2+2.3+2.2+3.4+3.2+...+n\left(n+1\right)+n.2\)

 \(\Rightarrow C=\left(1.2+2.3+3.4+...+n\left(n+1\right)\right)+2\left(1+2+3+...+n\right)\)

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

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

~ 

C = 1.4+2.5+3.6+4.7+...+n.(n+3)

C= 1.(2+2) + 2.(3+2) + 3.(4+2) + 4.(5+2) + ...+n.[(n+1) + 2]

C = 1.2 + 1.2 + 2.3 + 2.2 + 3.4 + 3.2 + 4.5 + 4.2 + ...+ n.(n+1) + n.2

C = [(1.2+2.3+3.4+4.5+...+n.(n+1)] + ( 1.2+2.2+3.2+4.2+...+n.2)

Đặt A = 1.2 + 2.3 + 3.4+4.5 + ...+n.(n+1)

=>3A = 1.2.3+2.3.3+3.4.3+4.5.3+...+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 + 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)

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

Đặt B = 1.2+2.2+3.2+4.2 +...+n.2

B = 2.(1+2+3+4+...+n)

B = 2. [(1+n).n:2]

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

B = (1+n).n

Thay A;B vào C

có: \(C=\frac{n.\left(n+1\right).\left(n+2\right)}{3}+\left(1+n\right).n\)

Ta thấy: 1.4 = 1.(1 + 3)

2.5 = 2.(2 + 3)

3.6 = 3.(3 + 3)

4.7 = 4.(4 + 3)

…….

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

Vậy C = 1.2 + 2.1 + 2.3 + 2.2 + 3.4 + 2.3 + … + n(n + 1) +2n

C = 1.2 + 2 +2.3 + 4 + 3.4 + 6 + … + n(n + 1) + 2n

C = [1.2 +2.3 +3.4 + … + n(n + 1)] + (2 + 4 + 6 + … + 2n)

⇒ 3C = 3.[1.2 +2.3 +3.4 + … + n(n + 1)] + 3.(2 + 4 + 6 + … + 2n) 

3C = 1.2.3 + 2.3.3 + 3.4.3 + … + n(n + 1).3 + 3.(2 + 4 + 6 + … + 2n)

3C = n(n + 1)(n + 2) + 

⇒ C =  +  = 

Ta thấy: 1.4 = 1.(1 + 3)

2.5 = 2.(2 + 3)

3.6 = 3.(3 + 3)

4.7 = 4.(4 + 3)

…….

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

Vậy C = 1.2 + 2.1 + 2.3 + 2.2 + 3.4 + 2.3 + … + n(n + 1) +2n

C = 1.2 + 2 +2.3 + 4 + 3.4 + 6 + … + n(n + 1) + 2n

C = [1.2 +2.3 +3.4 + … + n(n + 1)] + (2 + 4 + 6 + … + 2n)

⇒ 3C = 3.[1.2 +2.3 +3.4 + … + n(n + 1)] + 3.(2 + 4 + 6 + … + 2n) 

3C = 1.2.3 + 2.3.3 + 3.4.3 + … + n(n + 1).3 + 3.(2 + 4 + 6 + … + 2n)

3C = n(n + 1)(n + 2) + 

⇒ C =  +  = 

 3C = 3.[1.2 +2.3 +3.4 + ... + n(n - 1)] + 3.(2 + 4 + 6 + ... + 2n)

                    = 1.2.3 + 2.3.3 + 3.4.3 + ... + n(n - 1).3 + 3.(2 + 4 + 6 + ... + 2n)

Nên C  =  n(n-1)(n+5):3

Ta thấy: 1.4 = 1.(1 + 3)

2.5 = 2.(2 + 3)

3.6 = 3.(3 + 3)

4.7 = 4.(4 + 3)

…….

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

Vậy C = 1.2 + 2.1 + 2.3 + 2.2 + 3.4 + 2.3 + … + n(n + 1) +2n

C = 1.2 + 2 +2.3 + 4 + 3.4 + 6 + … + n(n + 1) + 2n

C = [1.2 +2.3 +3.4 + … + n(n + 1)] + (2 + 4 + 6 + … + 2n)

⇒ 3C = 3.[1.2 +2.3 +3.4 + … + n(n + 1)] + 3.(2 + 4 + 6 + … + 2n) 

3C = 1.2.3 + 2.3.3 + 3.4.3 + … + n(n + 1).3 + 3.(2 + 4 + 6 + … + 2n)

3C = n(n + 1)(n + 2) + 

⇒ C =  +  = 

Tính S = 1.4 + 2.5 + 3.6 + 4.7 + … + n(n + 3)
Lời giải
Ta thấy: 1.4 = 1.(1 + 3)
2.5 = 2.(2 + 3) 
3.6 = 3.(3 + 3) 
4.7 = 4.(4 + 3)
…….
n(n + 3) = n(n + 1) + 2n
Vậy S = 1.2 + 2.1 + 2.3 + 2.2 + 3.4 + 2.3 + … + n(n + 1) +2n
= 1.2 + 2 +2.3 + 4 + 3.4 + 6 + … + n(n + 1) + 2n
= [1.2 +2.3 +3.4 + … + n(n + 1)] + (2 + 4 + 6 + … + 2n)
3S = 3.[1.2 +2.3 +3.4 + … + n(n + 1)] + 3.(2 + 4 + 6 + … + 2n) =
= 1.2.3 + 2.3.3 + 3.4.3 + … + n(n + 1).3 + 3.(2 + 4 + 6 + … + 2n) =
= n(n + 1)(n + 2) +

S