Câu hỏi của Bùi Châu Anh

Question

CMR : 1/4 + 1/16 + 1/36 + 1/64 + 1/100 + 1/144 + ... + 1/10000 < 1/2

Nguyễn Phương Uyên · Accepted Answer

$Đ\text{ặt }S=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+....+\frac{1}{10000}$$S=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}$$S=\frac{1}{2^2}\cdot\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)$Ta có :$\frac{1}{2^2}< \frac{1}{1\cdot2};\text{ }\frac{1}{3^2}< \frac{1}{2\cdot3};\text{ }...;\text{ }\frac{1}{50^2}< \frac{1}{49\cdot50}$$\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}$$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}$$=1-\frac{1}{50}$$\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+1=2$$\Rightarrow\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< \frac{1}{2^2}\cdot2$$\Rightarrow S< \frac{1}{2}$ (ĐPCM)Câu trả lời: $Đ\text{ặt }S=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+....+\frac{1}{10000}$$S=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}$$S=\frac{1}{2^2}\cdot\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)$Ta có :$\frac{1}{2^2}< \frac{1}{1\cdot2};\text{ }\frac{1}{3^2}< \frac{1}{2\cdot3};\text{ }...;\text{ }\frac{1}{50^2}< \frac{1}{49\cdot50}$$\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}$$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}$$=1-\frac{1}{50}$$\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+1=2$$\Rightarrow\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< \frac{1}{2^2}\cdot2$$\Rightarrow S< \frac{1}{2}$ (ĐPCM)

SKT Khan · Answer

Đặt $A=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+....+\frac{1}{10000}$$\Rightarrow A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{100^2}$$\Rightarrow4A=1+\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{50^2}$$\Rightarrow4A< 1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}$$\Rightarrow4A=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}$$\Rightarrow4A< 2-\frac{1}{50}< 2$$\Rightarrow4A< 2\Rightarrow A< \frac{2}{4}=\frac{1}{2}$=>a<1/2Câu trả lời: Đặt $A=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+....+\frac{1}{10000}$$\Rightarrow A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{100^2}$$\Rightarrow4A=1+\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{50^2}$$\Rightarrow4A< 1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}$$\Rightarrow4A=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}$$\Rightarrow4A< 2-\frac{1}{50}< 2$$\Rightarrow4A< 2\Rightarrow A< \frac{2}{4}=\frac{1}{2}$=>a<1/2

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