Question

[Nguyễn Vũ Minh Hiếu]

Chứng tỏ rằng : \(\frac{1}{101}\) +    \(\frac{1}{102}\)+ .... +    \(\frac{1}{299}\)+    \(\frac{1}{300}\)>    \(\frac{2}{3}\)

Answer 1

[Phạm Trà Giang]

Ta đặt \(A=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}\)

Vì \(\frac{1}{101}>\frac{1}{102}>...>\frac{1}{299}>\frac{1}{300}\)

\(\Rightarrow A=\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)+\left(\frac{1}{201}+\frac{1}{202}+...+\frac{1}{300}\right)\)

\(\Rightarrow A>\left(\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\right)+\left(\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}\right)\)

\(\Rightarrow A>\left(\frac{1}{200}\cdot100\right)+\left(\frac{1}{300}\cdot100\right)\)

\(\Rightarrow A>\frac{1}{2}+\frac{1}{3}\)

\(\Rightarrow A>\frac{5}{6}>\frac{2}{3}\)

\(\Rightarrow\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}>\frac{2}{3}\)

Answer 2

[︻̷̿┻̿═━დდDarknightდდ]

\(\frac{1}{101}+\frac{1}{102}+....+\frac{1}{200}\)>\(\frac{1}{200}+\frac{1}{200}+\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\)=\(\frac{1}{2}\)(có 200 c/s\(\frac{1}{200}\))

\(\frac{1}{201}+\frac{1}{202}+\frac{1}{203}+...+\frac{1}{300}\)>\(\frac{1}{300}+\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}\)=\(\frac{2}{3}\)(có 200 c/s \(\frac{1}{300}\))

Vậy \(\frac{1}{101}+\frac{1}{102+}+....+\frac{1}{300}\)>\(\frac{1}{2}+\frac{2}{3}=\frac{2}{3}\) Đpcm

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