tick đã tui mới làm cho

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

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

3A=(1.2.3-0.1.2)+(2.3.4-1.2.3)+...+[n(n+1)(n+2)-(n-1)n(n+1)]

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

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

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

Đặt A= 1.2+2.3 +.......+99.100

3A= 1.2.3+2.3.4+3.4.3 +......+ 99.100.3

3A= 1.2. (3 - 0) + 2.3.(4 - 1) +3.4. (5 - 2)....... . 99.100. (101 - 98)

3A = (1.2.3 + 2.3.4 + 3.4.5 +...... + 99.100.101) - (0.1.2 + 1.2.3 + 2.3.4 +.......+ 98.99.100)

3A = 99.100.101 - 0.1.2

3A = 999900 - 0

3A= 999900

A= 999900 : 3

A = 333300 

A=1.2+2.3+3.4+…+99.100

3A = 1.2.3 + 2.3.3 + ... + 99.100.3

3A = 1.2.3 + 2.3.(4-1) + ... + 99.100.(101-98)

3A = 1.2.3 + 2.3.4 - 1.2.3 + ... + 99.100.101 - 98.99.100

3A = 99.100.101

=> A = \(\frac{99.100.101}{3}\)= 333 300

3A = 1 × 2 × 3 + 2 × 3 × ( 4 - 1 ) + ... + 99 × 100 × ( 101 - 98 )

3A = 1 × 2 × 3 + 2 × 3 × 4 - 1 × 2 × 3 + ... + 99 × 100 × 101 - 98 × 99 × 100

3A = 99 × 100 × 101 = 999900

A = 999900 ÷ 3 = 333300

Tích mình cái nha

ta thấy mỗi hạng tử của tổng trên là tích của hai số tự nhiên liên tiếp , khi đó:

gọi a₁=1.2=>3a₁=1.2.3=>3a₁=1.2.3-0.1.2

a₂=2.3=>3a₂=2.3.3=>3a₂=2.3.4-1.2.3

a₃=3.4=>3a₃=3.3.4=>3a₃=3.4.5-2.3.4

 .......

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

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

cộng các vế đẳng thức trên ta có:

3a₁+3a₂+...+3a_n-1+3a_n=1.2.3-0.1.2+2.3.4-1.2.3+...+(n-1)n(n+1)-(n-2)(n-1)n+n(n+1)(n+2)-(n-1)n(n+1)

=>3(a₁+a₂+...+a_n-1+a_n)=n(n+1)(n+2)

mà A=a₁+a₂+...+a_n-1+a_n nên 

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

=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + .... + 49.50.3

=> 3A = 1.2.3 + 2.3.( 4 - 1 ) + 3.4.( 5 - 2 ) + .... + 49.50.( 51 - 48 )

=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 49.50.51 - 48.49.50

=> 3A = ( 1.2.3 - 1.2.3 ) + ( 2.3.4 - 2.3.4 ) + .... + ( 48.49.50 - 48.49.50 ) + 49.50.51

=> 3A = 49.50.51

=> A = ( 49.50.51 ) : 3 

=> A = 41650

A = 1.2 + 2.3 + 3.4 + ..... + 49.50

3A=1.2.3+2.3.3+3.4.3+...+49.50.3

3A=1.2.3+2.3.(4-1)+3.4.(5-2)+...+48.49.(50-47)+49.50.(51-48)

3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+48.49.50-47.48.49+49.50.51-48.49.50

3A=(1.2.3-1.2.3)+(2.3.4-2.3.4)+...(47.48.49-47.48.49)-(48.49.50-48.49.50)+49.50.51

3A=0+0+...+0+0+49.50.51

3A=49.50.51

A=\(\frac{49.50.51}{3}\)

A=41650

Đáp số: A=41650

= 333 300

Đặt A = 1.2 + 2.3 + ... + 99.100

3A = 1.2.3 + 2.3.( 4 - 1 ) +....+ 99.100 . ( 101 - 98 )

3A = 1.2.3 + 2.3.4 - 1.2.3 +....+ 99.100.101 - 98.99.100

3A = 99 . 100 . 101

A = 999900 / 3

A = 333300

mo di em a.cach lam ma ngu thi tick bat cong thoi.ngo nhu bu

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

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

3S = 1.2.3 + 2.3.(4-1) + ...... + n.(n + 1).[(n + 2) - (n - 1)]

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

3S = (1.2.3 - 1.2.3) + (2.3.4 - 2.3.4)  +...... + [(n-1)n(n + 1) - (n - 1).n.(n + 1)] + n.(n + 1)(n + 2)

VẬy 3S = n.(n + 1)(n + 2)

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

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

3S=1.2.3+2.3.(4-1)+...+n.(n+1)(n+2-(n-1))

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

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

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

D = 1.2 + 2.3+ 3.4 +...+ 99.100

=>3D=1.2.3+2.3.3+3.4.3+...+99.100.3

=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+....+99.100.(101-98)

=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100

=99.100.101-0.1.2

=99.100.101

=999900

=>D=999900:3=333300

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

=>3Dn=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)]

=1.2.3-0.1.2+2.3.4-1.2.3+2.3.4-2.3.4+....+n(n+1)(n+2)-(n-1)n(n+1)

=n.(n+1).(n+2)-0.1.2

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

=>Dn=n.(n+1)(n+2):3

 =>điều cần chứng minh