Ta có
200-(3+2/3+...+2/100)
=200-(3+2(1/3+...+1/100)
=200-(3+2 (1-2/3+1-3/4+...+1-99/100))
=200-(3+2(98-(2/3+3/4+...+99/100)))
=200-3-196-(2/3+3/4+...+99/100)
=1-(2/3+3/4+...+99/100)
Thay:1-(2/3+3/4+...+99/100)/2/3+3/4+......+99/100=1/(1/2)=2
Chứng minh rằng: \(\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+\frac{2}{5}+..+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}}=2\)
Chứng tỏ rằng
[200-(3+2/3+2/4+2/5+...+2/100]:[1/2+2/3+3/4+...+99/100]=2
Chứng minh rằng 1/3-3/2^2+3/3^3-4/3^4+...+99/3^99-100/3^100
Chứng minh rằng 100- ( 1 + 1/2 +1/3 +...+1/100) = 1/2 +2/3 +3/4 +...+99/100
chứng minh rằng 100-(1+1/2+1/3+...+1/100)=1/2+2/3+3/4+...+99/100
Chứng minh rằng : 100 - (1 + 1/2 + 1/3 +...+1/100)= 1/2 +2/3 +3/4 +...+ 99/100
Chứng minh rằng
A=1/2 - 2/2^2 + 3/2^3 - 4/2^4 + ...99/2^99 - 100/2^100 < 2/9
Chứng minh rằng:
100-(1+1/2+1/3+...+1/100)=1/2+2/3+3/4+...+99/100
Chứng minh rằng : 1/3-2/3^2+3/3^3-4/3^4+...+99/3^99-100/3^100<3/16