Số số hạng là ( 100 - 1 ) : 1 + 1 = 100 số

Số cặp là   100 : 2 = 50 cặp 

Tổng mỗi cặp là  100 + 1 = 101

Tổng của dãy là  50 x 101 = 5050

100-99+98-97+96-95+...-3+2-1

= (100-99)+(98-97)+(96-95)+...+(2-1)

=1+1+1+..+1

=50 

(100+98+96+...+4+2)-(99+97+95+...+3+1)

=100+98+96+...+4+2-99-97-95-...-3-1

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

=1+1+...+1

=1 x 50

= 50

  (100 + 98 + 96 + .. + 4 + 2 ) - ( 99 + 97 + 95 + ... + 3 + 1 

= 100 + 98 + 96 +.. + 4 + 2  - 99  - 97 - 95 - ... - 3 - 1

= (1 00 - 99) + (98 - 97) +.. + (2 - 1 )

= 1 + 1 + 1 + .. + 1  (50 số 1)

= 1.50 

= 50

 ( 100 + 98 + 96 + .......... + 4 + 2 ) - ( 97 + 95 + 93 +...... + 1 )

= [ ( 100+ 2 ) x 50 : 2 ] - [ ( 97 + 1 ) x 49 : 2 ]

= 2550 - 2401

= 149

100+99-98-97+96+95-94-93+....+4+3-2-1

=(100+99-98-97)+(96+95-94-93)+....+(4+3-2-1)

=4+4+....+4(25 số hạng)

=4.25=100

  100 + 99  - 98 - 97 + ......... + 4 + 3 - 2 -1

= ( 100 + 99 - 98 - 97 ) + ........+ ( 4 + 3 - 2 -1 ) ( 25 nhóm )

= 4 + 4 + ....... + 4 ( 25 số 4 )

= 4 .25

 = 100

a)

C = 1 − 2 + 3 − 4 + ... + 97 − 98 + 99 − 100 = 1 − 2 + 3 − 4 + ... + 97 − 98 + 99 − 100 = − 1 + − 1 + ... + − 1 + − 1 = − 1.50 = − 50.

b)

B = 1 − 2 − 3 + 4 + 5 − 6 − 7 + ... + 97 − 98 − 99 + 100 = 1 − 2 + − 3 + 4 + 5 − 6 + ... + 97 − 98 + − 99 + 100 = − 1 + 1 + − 1 + ... + − 1 + 1 = − 1 + 1 + − 1 + 1 + ... + − 1 + 1 − 1 = 0 + 0 + ... + 0 − 1 = − 1.

Ta có: \(M=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(1+\dfrac{2}{98}\right)+\left(1+\dfrac{3}{97}\right)+\left(1+\dfrac{4}{96}\right)+...+\left(1+\dfrac{98}{2}\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+\dfrac{100}{1}+\dfrac{100}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

=100

Ta có: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)

\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{8}{\dfrac{1}{5}}=40\)

\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)

A= 1+1+1+1+1..........+1

A có số số 1 là 

(100-2):2 +1= 50

tổng đó là

(100+2).50:2=2550