Ta có:\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
Tính A=1/51+1/52+1/53+...+1/99+1/100
tính 1/51+1/52+1/53+....+1/100
1/1*2+1/3*4+1/5*6+...+1/99*100
tính c=1/1*2+1/3*4+1/5*6+...+1/99*100
tính b=1/51+1/52+1/53+...+1/100
1/51+1/52+1/53+......+1/99+1/100 >1/2
(1/51+1/52+1/53+...+1/100)÷(1/1×2+1/3×4+1/4×5+...+1/99×100)
1/51+1/52+1/53+...1/99+1/100 hay so sanh voi 1/2
A = 1 . 2 + 2 . 3 + 3 . 4 + ......... + 98 . 99 / 1 + ( 1 + 2 ) + ( 1 + 2 + 3 ) + ........... + ( 1 + 2 + 3 + ...... + 98 )
B = ( 1 / 51 . 52 ) + 1 / 52 . 53 + ...... + 1 / 100 . 101 ) : ( 1 / 1 . 2 + 1 / 2 . 3 + ........ + 1 / 99 . 100 + 1 / 100 . 101
Chứng tỏ A = 1/51 + 1/52 + 1/53 + .....+1/99 + 1/100 <1/2
CMR(1/1*2+1/2*3+1/3*4+1/4*5+...+1/99*100):(1/51+1/52+1/53+...+1/100) = 1
