\(\dfrac{2}{1\times4}+\dfrac{2}{4\times7}+\dfrac{2}{7\times10}+\cdot\cdot\cdot+\dfrac{2}{97\times100}\)

\(=\dfrac{2}{3}\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\cdot\cdot\cdot+\dfrac{3}{97\times100}\right)\)

\(=\dfrac{2}{3}\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\cdot\cdot\cdot+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

\(=\dfrac{2}{3}\times\left(1-\dfrac{1}{100}\right)\)

\(=\dfrac{2}{3}\times\dfrac{99}{100}\)

\(=\dfrac{33}{50}\)

#\(Toru\)

A=2/3 x (1-1/4+1/4-1/7+......+1/97-1/100)

  = 2/3 x (1-1/100)

  = 2/3 x 99/100

  = 33/50

A=\(\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+........+\frac{3}{97.100}\right)\)

=\(\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...........+\frac{1}{97}-\frac{1}{100}\right)\)

=\(\frac{2}{3}.\left(1-\frac{1}{100}\right)\)

=\(\frac{2}{3}.\frac{99}{100}\)

=\(\frac{33}{50}\)

33/50

\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)

= \(\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

= \(\frac{2}{3}\left(1-\frac{1}{100}\right)\)

= \(\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)

\(D=\frac{2}{1.4}+\frac{2}{4.7}+...+\frac{2}{97.100}\)

\(3D=\frac{2.3}{1.4}+\frac{2.3}{4.7}+...+\frac{2.3}{97.100}\)

\(3D=2\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)\)

\(3D=2\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(3D=2\left(1-\frac{1}{100}\right)\)

\(3D=2\cdot\frac{99}{100}\)

\(3D=\frac{99}{50}\)

\(D=\frac{99}{50}:3\)

\(D=\frac{33}{50}\)

B = \(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)

= \(\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

= \(\frac{2}{3}\left(1-\frac{1}{100}\right)=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)

\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+...+\frac{3^2}{97.100}\)

\(A=\frac{3^2}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{97}-\frac{1}{100}\right)\)

\(A=3\cdot\left(1-\frac{1}{100}\right)\)

\(A=3\cdot\frac{99}{100}=\frac{297}{100}\)

Vậy \(A=\frac{297}{100}\)

A = 1 - 2 - 3 - 4 + 5 - 6 -  7 - 8 + ........... + 97 - 98 - 99 - 100 (100 số )

A = (1 - 2 - 3 - 4) + (5 - 6 - 7 - 8) + ................ + (97 - 98 - 99 - 100)

(25 cặp , tính bằng cách lấy số cả dãy chia cho số số của mỗi cặp )

A = (-8) . 25 

A = -200

59 nha

A = 1-2 + 3-4 + ......+ 99-100

= ( -1) + ( -1) +....+(-1) 

= (-1) x 50 

= -50

vì có tấ cả 50 số -1

A = 1 + 3 + 5 + ... + 101

A = ( 101 + 1) x 51 : 2

A = 2061

B = 1 + 4 + 7 + 10 + ...+ 100

B = ( 1 + 100) x 34 :2

B = 1717

E=1+2-3-4+5+6-7-8+...-299-300+301+302

=1+(2-3-4)+5+(6-7-8)+...+(298-299-300)+301+302

=1-5+5-9+...-301+301+302

=1+302=303

C = 100 - 99 + 98 - 97 + ...+ 2 -1

C = ( 100 + 98 + ...+ 2) - (99+97+...+1)

C = (2+100) x 50 : 2 - (1+99) x 50 : 2

C = 2550- 2500

C = 50

E = 1 + 2 - 3 -4 + 5 + 6 -  7 - 8 + ...-299 -300 + 301 + 302

E = 1 + ( 2-3-4+5) + (6-7-8+9) + ...+ ( 298-299-300+301) + 302

E = 1 + 0 + 0 + ...+ 0 + 302

E = 303