Question

Cho\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

Chứng minh rằng\(\frac{7}{12}\)< A < \(\frac{5}{6}\)

Answer 1

Ta thấy : A = 1 - 1/2 + 1/3 - 1/4 +...+ 1/99 +1/100

             A = 1 + 1/2 + 1/3 + 1/4 +...+ 1/99 + 1/100 - 2. (1/2 + 1/4 +1/6 +...+ 1/100)

            A = 1 + 1/2 + 1/3 +1/4 +...+ 1/99 + 1/100 - (1 + 1/2 + 1/3 +...+1/50)

            A = 1/51 + 1/52 + 1/53 +...+ 1/100

Do đó : A = (1/51 + 1/52 + 1/53 +...+ 1/75) + (1/76 + 1/77 +...+ 1/100)

Ta có : 1/51 > 1/52 > ... > 1/75    ;     1/76 > 1/77 > ... > 1/100 nên :

           A > 1/75.25 + 1/100.25 = 1/3 + 1/4 = 7/12

           A < 1/51.25 < 1/50.25 + 1/75.25 = 1/2 +1/3 = 5/6

Vậy 7/12 < A <5/6 . ^_^