​A rê. Lớp 6 ngược mà hỏi bài đó hở

đây là bài của bảo trân

Nguyễn Huy Thắng hok lớp 9 ak

Đặt A  = ( 1² + 2² + 3^{^2} + ... + 49² ) - ( 1.2 + 2.3 + ....+ 49.50 ) 

Đặt S = 1² + 2² + 3² +...+ 49²

S = (1.2 - 1) + (2.3 - 2) + (3.4 - 3) +...+ (49.50 - 49)

S = (1.2 + 2.3 + 3.4 +...+ 49.50) - (1 + 2 + 3 +...+ 49)

=> A = [(1.2 + 2.3 + 3.4 +...+ 49.50) - (1 + 2 + 3 +...+ 49)] - (1.2 + 2.3 +...+ 49.50)

A = [(1.2 + 2.3 + 3.4 +...+ 49.50) - (1.2 + 2.3 +...+ 49.50)] - (49 + 1)49 : 2  (công thức tính tổng)

A = 0 - 1225

A = -1225

Mình tính ra bằng -1225
không biết có đúng không ? 

gvsg574679jnxcnkvfr9684589643-95lkmhxk;lmcb

\(B=1.2+2.3+3.4+...+49.50\)

\(\Rightarrow3B=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)

\(\Rightarrow3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)

\(\Rightarrow3B=49.50.51\)

\(\Rightarrow B=\frac{49.50.51}{3}\)

B=\(1.2+2.3+....+49.50\\ \Rightarrow3B=1.2.\left(3-0\right)+2.3.\left(4-1\right)+.....+49.50.\left(51-48\right)\)

\(\Rightarrow3B=1.2.3-0.1.2+2.3.4-1.2.3.+.....+49.50.51-48.49.50\\ \Rightarrow3B=49.50.51\\ \Rightarrow3B=124950\\ \Rightarrow B=41650\)

C=\(1^2+2^2+3^2+....+50^2\\ =1.1+2.2+3.3+.....+50.50\\ =1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+....+50\left(51-1\right)\\ \)

\(=\left(1.2+2.3+3.4+.....+50.51\right)-\left(1+2+3+....+50\right)\)

Áp dụng bài trên để tính 

thế mà chả đứa nào trả lời câu hỏi của mk

3N = 1.2.3+2.3(4-1)+3.4.(5-2)+.+99.100.(101-98)

3N = 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+.+99.100.101-98.99.100

3N = 99.100.101

3N=33.100.101=333300

b)

tổng này có  99-10+1=90 (số hạng):

10,11 + 11,12 + 12,13 +............+ 98,99 + 99,100 =

10,100 + 11,11 + 12,12 + .......... + 98,98 + 99,99 =

(10,10 + 99,99) x 90 : 2 = 4954,05

c)

R=1.(2-1)+2.(3-1)+.....+100.(101-1)

=1.2-1.1+2.3-1.2+......+100.101-1.100

=(1.2+2.3+.....+99.100+100.101)-(1+2+3+...+100)

=[1.2.3+2.3.(4-1)+........100.101.(102-99)]:3+[(100+1).100:2]

(tổng trên chia cho 3 nên cuối cùng chia 3)

=(1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+.....100.101.102-99.100.101):3+5050

=(100.101.102) :3 +5050

=348450

d)=1.100+2.(100-1)+.....+100.(100-99)

=1.100+2.100-1.2+3.100-2.3+........+100.100-99.100

=100.(1+2+3+.......+100)-(1.2+2.3+3.4+....+99.100)

=100.\(\frac{101.100}{2}-\frac{99.100.101}{3}\) =505000-333300=171700

p/s mỏi tay, bấm mình nhé

a)N=1.2+2.3+3.4+4.5+......+99.100

3N=1.2.3+2.3.3+3.3.4+3.4.5+......+3.99.100

3N=1.2.3+2.3.(4-1)+3.4.(5-2)+...........+99.100.(101-98)

3N=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+.......+99.100.101-98.99.100

3N=99.100.101=999900

N=333300

b)P=10.11+11.12+12.13+....+29.30

3P=10.11.3+11.12.3+....+29.30.3

3P=10.11.12-9.10.11+11.12.13-10.11.12+.........+29.30.31-28.29.30

3P=29.30.31-9.10.11=25980

P=8660

R=1²+2²+3²+....+100²

R=1.(2-1)+2.(3-1)+.........+100.(101-1)

R=1.2-1+2.3-2+3.4-3+.....+100.101-100

R=(1.2+2.3+3.4+..........+100.101)-(1+2+3+.........+99+100)

\(R=\left(N+100.101\right)-\frac{100.101}{2}\)

R=343400-5050=338350

S=100.1+2.(100-1)+3.(100-2)+......+99.(100-98)+100(100-99)

S=100.1+100.2-1.2+100.3-2.3+.......+99.100-98.99+100.100-99.100

S=100(1+2+...+100)-(1.2+2.3+....+99.100)

S=100.(100.101:2)-N

S=505000-333300=171700

x2 là \(x^2\)hả bạn

a) x²+20x+*

=> x² +2 x 5x²+5²

= (x+5)²

b) 16x²+24xy+*

=> (4x)²+2 x 4x  x  3+3²

= (4x + 3)²

c) y² -*+49

=> y² - 2y7²+7²

= (y-7)²

d) * -  42xy + 49y²

= (3x)² + 2 x 7y3x + (7y)²

= (3x+7y)²

a) x² + 20x + *

= x² + 2.x.10 + 10²

= x² + 20x + 100

b) 16x² + 24xy +*

= 16x² +2.x.12y + (12y)²

⁼ 16x² +24xy + 144y²

c) y² - * + 49

= y² - * +7²

= y² - 2.y.7 + 49

= y² - 14y + 49

d) * - 42xy + 49y²

= * - 42xy + (7y)²

⁼ * - 2.3x.7y + (7y)²

= (3x)² - 42xy + (7y)²

= 9x² - 42xy + 49y²

a, 1 + 2 + 3 + ... + n = \(\left[\frac{n-1}{1}+1\right]\left[n+1\right]\)

1 + 3 + 5 + 7 + ... + [2n-1] = \(\left[\frac{2n-1-1}{2}+1\right]\left[2n-1+1\right]\)

b, A = 1.2+2.3+3.4+...+n[n+1] 

=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + n[n+1].3

Mà: 1.2.3 = 1.2.3 - 0.1.2

       2.3.3 = 2.3.4 - 1.2.3 

  .......................................

      n[n+1].3 = n[n+1][n+2] - [n-1]n[n+1]

=> 3A = [n-1]n[n+1]

=> A = \(\frac{\left[n-1\right]n\left[n+1\right]}{3}\)

1.2.3.+2.3.4+...+n[n+1][n+2]

4A = 1.2.3.[4-0] + 2.3.4.[5-1] + .... + n[n+1][n+2].[[n+3] - [n-1]]

4A =  1.2.3.4 - 0.1.2.3 + 2.3.4.5 - 1.2.3.4 +...+ n[n+1][n+2][n+3] - n[n+1][n+2][n-1]

4A = 1.2.3.4 - 1.2.3.4 + 2.3.4. 5 - 2.3.4.5 + ... + n[n+1][n+2][n+3] - n[n+1][n+2][n+3] + n[n+1][n+2][n-1]

4A = n[n+1][n+2][n-1]

A = \(\frac{\text{n[n+1][n+2][n-1]}}{4}\)

sai đề 

sai đề là cái chắc

sai đề hả bạn????