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Chung minh rang S=1/5^2+1/6^2+1/7^2+....+1/100^2 nho hon 1/2

Le Thi Khanh Huyen · Accepted Answer

Ta có:$\frac{1}{5^2}<\frac{1}{4.5}$$\frac{1}{6^2}<\frac{1}{5.6}$$...$$\frac{1}{100^2}<\frac{1}{99.100}$$\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}<\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}$$\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}<\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}$$\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}<\frac{1}{4}-\frac{1}{100}<\frac{1}{4}<\frac{1}{2}$Vậy $\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}<\frac{1}{2}$Câu trả lời: Ta có:$\frac{1}{5^2}<\frac{1}{4.5}$$\frac{1}{6^2}<\frac{1}{5.6}$$...$$\frac{1}{100^2}<\frac{1}{99.100}$$\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}<\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}$$\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}<\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}$$\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}<\frac{1}{4}-\frac{1}{100}<\frac{1}{4}<\frac{1}{2}$Vậy $\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}<\frac{1}{2}$

Nguyên · Answer

$s=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}$$S=\frac{1}{5.5}+\frac{1}{6.6}+\frac{1}{7.7}+...+\frac{1}{100.100}<\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}$$S<\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{101}$$\Rightarrow S<\frac{1}{5}-\frac{1}{101}$Vì $\frac{1}{5}<\frac{1}{2}$nên $\frac{1}{5}-\frac{1}{101}<\frac{1}{2}$hay $S=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}<\frac{1}{5}-\frac{1}{101}<\frac{1}{2}$Vậy $S=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}<\frac{1}{2}$  (đpcm)Câu trả lời: $s=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}$$S=\frac{1}{5.5}+\frac{1}{6.6}+\frac{1}{7.7}+...+\frac{1}{100.100}<\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}$$S<\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{101}$$\Rightarrow S<\frac{1}{5}-\frac{1}{101}$Vì $\frac{1}{5}<\frac{1}{2}$nên $\frac{1}{5}-\frac{1}{101}<\frac{1}{2}$hay $S=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}<\frac{1}{5}-\frac{1}{101}<\frac{1}{2}$Vậy $S=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}<\frac{1}{2}$  (đpcm)

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