Câu hỏi của Nguyễn Dạ Thảo An

Question

Cho A = 1/2^2+1/3^2+1/4^2+...+1/48^2+1/49^2+1/50^2.Chứng tỏ A<1

I don't need you · Accepted Answer

ta có: $\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{48^2}< \frac{1}{47.48};\frac{1}{49^2}< \frac{1}{48.49};\frac{1}{50^2}< \frac{1}{49.50}$$\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{48^2}+\frac{1}{49^2}+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{47.48}+\frac{1}{48.49}+\frac{1}{49.50}$                                                                                               $=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{47}-\frac{1}{48}+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}$$=1-\frac{1}{50}=\frac{49}{50}< 1$$\Rightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{48^2}+\frac{1}{49^2}+\frac{1}{50^2}< 1\left(đpcm\right)$Câu trả lời: ta có: $\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{48^2}< \frac{1}{47.48};\frac{1}{49^2}< \frac{1}{48.49};\frac{1}{50^2}< \frac{1}{49.50}$$\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{48^2}+\frac{1}{49^2}+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{47.48}+\frac{1}{48.49}+\frac{1}{49.50}$                                                                                               $=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{47}-\frac{1}{48}+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}$$=1-\frac{1}{50}=\frac{49}{50}< 1$$\Rightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{48^2}+\frac{1}{49^2}+\frac{1}{50^2}< 1\left(đpcm\right)$

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