Question

\(Cho:A=\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{80}\)

Chứng minh rằng : \(A>\dfrac{7}{12}\)

Answer 1

Ta có:

7/12 = 4/12 + 3/12 = 1/3 + 1/4 = 20/60 + 20/80

1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 = (1/41 + 1/42 + 1/43 + ...+ 1/60) + (1/61 + 1/62 +...+ 1/79 + 1/80)

Do 1/41> 1/42 > 1/43 > ...>1/59 > 1/60 => (1/41 + 1/42 + 1/43 + ...+ 1/60) > 1/60 + ...+ 1/60 = 20/60 và 1/61> 1/62> ... >1/79> 1/80 => (1/61 + 1/62 +...+ 1/79 + 1/80) > 1/80 + ...+ 1/80 = 20/80

Vậy: 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 20/60 + 20/80 = 7/12 => 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 7/12 => ĐPCM

Answer 2

Ta có : 1/41 + 1/42 + ... + 1/60 > 1/60 * 20 = 1/3 .

1/61 + 1/62 + ... + 1/80 > 1/80 * 20 = 1/4 .

⇒ 1/41 + 1/42 + ... + 1/80 > 1/3 + 1/4 = 4/12 + 3/12 .

= 7/12 .

Do đó : A > 7/12 .

Vậy bài toán được chứng minh .