a) B= 1 + 3 + 7 + 9 +...+ 99

= ( 1 + 99 ) + ( 97 + 3 ) + ( 95 + 5 ) + ( 93 + 7 ) + ( 91 + 9 ) + ( 89 + 11 ) + ... ,

(Tổng cộng có 25 cặp có tổng là 100.)
= 25 x 100.
B = 2500.

b,    C = 1+4+7+10+..............+100

Số số hạng là : 

(100-1) : 3 + 1 = 34 (số)

C = (100+1) x 34 : 2 

C = 101 x 34  : 2

C = 3434 : 2

C = 1717

1/2 + 1/3 + 1/4 + 1/5 + 100

= 30/60 + 20/60 + 15/60 + 12/60 + 100

= 77/60 +100

= 385/300 + 300/300

= 685/300

= 137/100

`A=1/2.3+1/3.4+...+1/99.100`

`=1/2-1/3+1/3-1/4+...+1/99-1/100`

`=1/2-1/100`

`=50/100-1/100`

`=49/100`

-1+2-3+4-5+6-...-2021+2022

=(-1+2)+(-3+4)+...+(-2021+2022)

=1+1+1+...+1 +1+1 (vì ở đây là 1011 số nhé cậu)
=1011

\(=-1-1-1-...-1=-1\cdot10=-10\)

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

Ta có : \(M=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+.....+\frac{4}{95.99}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+......+\frac{1}{95}-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}\)

\(=\frac{32}{99}\)

Ta có ; K = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)

\(=1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{90}\)

\(=1+\left(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{9.10}\right)\)

\(=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}\right)\)

\(=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{9}-\frac{1}{10}\right)\)

\(=1+2\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(=1+1-\frac{1}{5}\)(nhân phá ngoặc)

\(=2-\frac{1}{5}\)< 2 

Vậy K = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)< 2

3.A=\(\frac{1}{3^2}+\frac{1}{3^3}+......\frac{1}{3^{101}}\)

3.A-A=\(\frac{1}{3^{101}}-1\)

2.A=1

A=1/2