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Question

Chứng minh rằng :$\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}< 1\
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Minh Tâm · Accepted Answer

Đặt A=$\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{2010^2}$Ta có: $\frac{1}{2^2}< \frac{1}{1\cdot2};\frac{1}{3^2}< \frac{1}{2\cdot3};\frac{1}{4^2}< \frac{1}{3\cdot4};....;\frac{1}{2010^2}< \frac{1}{2009\cdot2010}$=> A<$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2009\cdot2010}$$\Leftrightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2009}-\frac{2}{2010}$$\Leftrightarrow A< 1-\frac{1}{2010}$<=> A<1 (đpcm)Câu trả lời: Đặt A=$\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{2010^2}$Ta có: $\frac{1}{2^2}< \frac{1}{1\cdot2};\frac{1}{3^2}< \frac{1}{2\cdot3};\frac{1}{4^2}< \frac{1}{3\cdot4};....;\frac{1}{2010^2}< \frac{1}{2009\cdot2010}$=> A<$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2009\cdot2010}$$\Leftrightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2009}-\frac{2}{2010}$$\Leftrightarrow A< 1-\frac{1}{2010}$<=> A<1 (đpcm)

Lê Thị Nhung · Answer

Ta có $\frac{1}{2^2}< \frac{1}{1.2}$$\frac{1}{4^2}< \frac{1}{3.4}$$\frac{1}{3^2}< \frac{1}{2.3}$$\frac{1}{4^2}< \frac{1}{3.4}$...$\frac{1}{2010^2}< \frac{1}{2009.2010}$Cộng vế các BĐT trên ta được$\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}$$\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}$$\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}< 1-\frac{1}{2010}< 1$Vậy $\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}< 1$Câu trả lời: Ta có $\frac{1}{2^2}< \frac{1}{1.2}$$\frac{1}{4^2}< \frac{1}{3.4}$$\frac{1}{3^2}< \frac{1}{2.3}$$\frac{1}{4^2}< \frac{1}{3.4}$...$\frac{1}{2010^2}< \frac{1}{2009.2010}$Cộng vế các BĐT trên ta được$\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}$$\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}$$\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}< 1-\frac{1}{2010}< 1$Vậy $\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}< 1$

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