a,

A=1−3−5−7−9−...−97−99a)A=1−3−5−7−9−...−97−99 

=1−(3+5+7+...+99)=1−(3+5+7+...+99)

=1−(99+3).[(99−3):2+1]2=1−(99+3).[(99−3):2+1]2
=1−2499=−2498=1−2499=−2498

b)B=1+3−5−7+9+...+97−99b)B=1+3−5−7+9+...+97−99
=(−8)+(−8)+(−8)+...+(−8)+97−99=(−8)+(−8)+(−8)+...+(−8)+97−99
=(−8).12+(−2)=−98=(−8).12+(−2)=−98

c)C=1−3−5+7+9−11−13+15+...+97−99c)C=1−3−5+7+9−11−13+15+...+97−99
=0+0+0+0+0+...+0−99=0+0+0+0+0+...+0−99
=−99

`A=1/(1xx3)+1/(3xx5)+1/(5xx7)+...+1/(95xx97)+1/(97xx99)`

`=> 2A=2/(1xx3)+2/(3xx5)+...+2/(97xx99)`

`=> 2A=1-1/3+1/3-1/5+...+1/97-1/99`

`=> 2A=1-1/99`

`=> 2A=98/99`

`=> A=49/99`

cái này bạn mở sách bồi dưỡng toán ra trang gần cuối là thấy ngay ấy mà

C=\(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

  =\(\frac{1}{100}-\left(\frac{1}{2.1}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

  =\(\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

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

  =\(\frac{1}{100}-\frac{99}{100}\)

  =\(\frac{-98}{100}=\frac{-49}{50}\)

C=1/100 -1/100.99 -1/99.98 -1/98.97-......- 1/3.2 -1/2.1 
= 1/100 - (1/100.99 + 1/99.98 + 1/98.97-......+ 1/3.2 +1/2.1) 
Đặt A = 1/100.99 + 1/99.98 + 1/98.97-......+ 1/3.2 +1/2.1 => C = 1/100 - A 
Dễ thấy 1/2.1 = 1/1 - 1/2 
1/3.2 = 1/2 - 1/3 
..................... 
1/99.98 = 1/98 - 1/99 
1/100.99 = 1/99 - 1/100 
=> cộng từng vế với vế ta

\(B=\left(\frac{3}{7}+\frac{-3}{7}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{5}{9}+\frac{-5}{9}\right)+\left(\frac{2}{11}-\frac{2}{11}\right)\)

\(+\left(\frac{7}{13}-\frac{7}{13}\right)-\frac{9}{16}\)

\(=0+0+0+0-\frac{1}{16}\)

\(=\frac{-1}{16}\)

Ta có:

\(C= 4+44+444+......+4444444444\)

\(C= 4.(10.1+9.10+8.100+7.1000+...+1.1000000000\)

\(C= 4.(100+90+800+7000+60000+500000+4000000+30000000+200000000+1000000000)\)

\(C=4.12345678900\)

\(C=4938271600\)

Tương tự.