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

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

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

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

\(=\dfrac{n\left(n+1\right)\left(2n+1\right)+9n\left(n+1\right)}{6}\)

\(=\dfrac{n\left(n+1\right)\left(2n+1+9\right)}{6}\)

\(=\dfrac{n\left(n+1\right)\left(2n+10\right)}{6}=\dfrac{n\left(n+1\right)\left(n+5\right)}{3}\)

22 tháng 5 2021

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) + \frac{3\left(2n\ +\ 2\right)n}{2}

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

22 tháng 5 2021

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) + \frac{3\left(2n\ +\ 2\right)n}{2}

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

13 tháng 11 2023

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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: \(A=\dfrac{n\left(n+1\right)}{2}\)

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

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

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

4 tháng 9 2016

\(A=1\left(2+2\right)+2\left(2+3\right)+3\left(2+4\right)+.....+\left(n-1\right)\left(2+n\right)\)

\(\Leftrightarrow A=1.2+1.2+2.3+2.2+3.4+2.3+....+\left(n-1\right)n+2\left(n-1\right)\)

\(\Leftrightarrow A=\left(1.2+2.3+.....+\left(n-1\right)n\right)+2\left(1+2+3+....+\left(n-1\right)\right)\)

Giả sử A=B+C

Với \(\begin{cases}B=1.2+2.3+.....+\left(n-1\right)n\\C=2\left[1+2+....+\left(n-1\right)\right]\end{cases}\)

Ta có

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

\(\Rightarrow3B=1.2.3-0.1.2+2.3.4-1.2.3+.....+\left(n-1\right)n\left(n+1\right)-\left(n-2\right)\left(n-1\right)n\)

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

Mặt khác

\(C=2\left[1+2+....+\left(n-1\right)\right]\)

\(\Rightarrow C=2.\frac{\left[\left(n-1\right)+1\right]n}{2}=n^2\)

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

Vậy \(A=\frac{\left(n-1\right)n\left(n+1\right)}{3}+n^2\)

28 tháng 11 2021

TK

S=1.4+2.5+3.6+4.7+....+n.(n+3) S = 1. ( 2 + 2 ) + 2. ( 3 + 2 ) + 3. ( 4 + 2 ) + . . . + n . [ ( n + 1 ) + 2 ] S = 1.2 + 2.3 + 3.4 + . . . . + n . ( n + 1 ) + ( 1.2 + 2.2 + 3.2 + . . . . + n .2 ) Đặt A = 1.2 + 2.3 + 3.4 + . . . . + n . ( n + 1 ) 3 A = 1.2.3 + 2.3. ( 4 − 1 ) + . . . . + n . ( n + 1 ) . [ ( n + 2 ) − ( n − 1 ) 3 A = 1.2.3 + 2.3.4 − 1.2.3 + . . . . + n . ( n + 1 ) . ( n + 2 ) − ( n − 1 ) . n . ( n + 1 ) 3 A = n . ( n + 1 ) . ( n + 2 ) A = [ n . ( n + 1 ) . ( n + 2 ) ] : 3 S = [ n . ( n + 1 ) . ( n + 2 ) ] : 3 + 2. ( 1 + 2 + 3 + . . . + n ) S = [ n . ( n + 1 ) . ( n + 2 ) ] : 3 + 2. n . ( n + 1 ) : 2 S = n . ( n + 1 ) . ( n + 2 ) : 3 + n . ( n + 1 ) S = n . ( n + 1 ) . [ ( n + 2 ) : 3 + 1 )

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

 


   = [ 1.2 + 2.3 + 3.4 + ..... + n.( n + 1 ) ] - ( 1 + 2 + 3 + .... + n ) 
   = { [ n.( n+1 ).( n+2 )] /3 } - { [ n.( n+1)] /2 } 
   = { n(n+1)(2n+1) }/ 6 
Vậy......... 

28 tháng 11 2021

TK

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20 tháng 9 2017

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

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

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


26 tháng 9 2017

Dựa theo công thức tự thiết kế do các anh em trong đoàn ( những con người ẩn danh ) là : { k . ( k + 3 ) = k . ( k + 1 ) + 2 . k }

Ta có :

S = 1 . 4 + 2 . 5 + 3 . 6 + . . . + n . ( n + 3 )

S = ( 1 . 2 + 2 . 1 ) + ( 2 . 3 + 2 . 2 ) + . . . + [ n . ( n + 1 ) + 2 . n ]

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

Dựa theo công thức số 37 và 55 quyển 7 của các em trong đoàn .

Ta có :

S = [ n . ( n + 1 ) . ( n + 2 ) ] + ( \(n^2\)+ n ) ]