vào phần câu hỏi tương tự nhé bn,trong đó có đấy

đạt A= 1.2 + 2.3 + 3.4 +.............+ 98.99

3A=1.2.3+2.3.3+.....+98.99.3

3A=1.2.3+2.3.4-2.3.1+.....+98.99.100-99.98.97

3A=98.99.100

A=98.33.100

A = 1.2+2.3+3.4+......+99.100 
Gấp A lên 3 lần ta có: 
A . 3 = 1.2.3 + 2.3.3 + 3.4.3 + … + 99.100.3 
A . 3 = 1.2.3 + 2.3.(4 - 1) + 3.4.( 5 - 2) + … + 99.100. (101 - 98) 
A . 3 = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + … + 99.100.101 - 98.99.100 
A . 3 = 99.100.101 
A = 99.100.101 : 3 
A = 33.100.101 
A = 333 300

Đặt tổng trên = A

Có : 3A = 1.2.3+2.3.3+....+98.99.3

 = 1.2.3+2.3.(4-1)+.....+98.99.(100-97)

 = 1.2.3+2.3.4-1.2.3+.....+98.99.100-97.98.99

 = 98.99.100

=> A = 98.99.100/3 = 323400

k mk nha

Gọi A = 1.2 + 2.3 + .. + 98.99

3A = 1.2.3 + 2.3.3 + ... + 98.99.3

3A = 1.2.3 + 2.3.(4 - 1) + ... + 98.99.(100 - 97)

3A = 1.2.3 + 2.3.4 - 1.2.3 + ... + 98.99.100 - 97.98.99

3A = 98.99.100

3A = 970200

A = 323400

B = 1.2 + 3.4 + 5.6 +...+ 98.99

3B = 1.2(3-0) + 2.3(4-1) ...... 98.99 (100-97)

3B = 1.2.3-0.2.3.4-1 ......... 98.99.100-97

3B = ( 1.2.3+2.3.4+.....+98.99.100) - ( 0.1.2+ 1.2.3+.... + 97.98.99)

3B = 98.99.100

3B = 970200

  B = 323400

C=1*2+2*3+3*4+...+98*99

C=2+6+12+...+9702

C=2+9702

C=9704

vay C=9704

D=(1*99+2*99+3*99+...+99*99)-(1*2+2*3+3*4+...+98*99)

D=(99+198+297+...+9801)-(2+6+12+...+9702)

D=(99+9801)-(2+9702)

D=9900-9704

D=196

vay D=196

ai di qua dong tinh thi nho h cho minh nhe

Ta có: \(A=1.2+2.3+...+98.99\)

\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+98.99.\left(100-97\right)\)

\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+98.99.100-97.98.99\)

\(\Rightarrow3A=98.99.100\)

\(\Rightarrow A=\frac{98.99.100}{3}\)

\(\Rightarrow A=98.33.100\)

\(\Rightarrow A=323400\)

A=1.2+2.3+3.4+........+98.99
3A=1.2.3+2.3.3+3.4.3+........+98.99.3
3A=1.2.3+2.3.(4 -1) +3.4.(5 -2)+........+98.99.(100 -97)
3A=1.2.3+2.3.4 -1.2.3 +3.4.5 -2.3.4 +........+98.99.100 -97.98.99
3A=98.99.100
===>A=(98.99.100)/3

\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{98.99}+\frac{2}{99.100}\)

= \(2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

= \(2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

= \(2\left(1-\frac{1}{100}\right)\)

 =\(2.\frac{99}{100}\)

 =\(\frac{99}{50}\)

Số các số hạng của A là :

( 99 - 1 ) : 1 + 1 = 99 ( số )

Tổng A là :

( 99 + 1 ) . 99 : 2 = 4950

Vậy tổng A = 4950 .

Bạn nhân 3 lên