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Question

Chứng minh rằng: $A=\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{49^2}+\frac{1}{50^2}<2$

BAN is VBN · Accepted Answer

Ta có: $\frac{1}{2^2}<\frac{1}{1\cdot2}$           $\frac{1}{3^2}<\frac{1}{2\cdot3}$           .....................           $\frac{1}{50^2}<\frac{1}{49\cdot50}$$\Rightarrow\frac{1}{2^2}+...+\frac{1}{49^2}+\frac{1}{50^2}<\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}$Ta có: $\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}$$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}$$=1-\frac{1}{50}<1$$\Rightarrow\frac{1}{2^2}+...+\frac{1}{49^2}+\frac{1}{50^2}<1$$\Rightarrow A=\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{49^2}+\frac{1}{50^2}<2$Câu trả lời: Ta có: $\frac{1}{2^2}<\frac{1}{1\cdot2}$           $\frac{1}{3^2}<\frac{1}{2\cdot3}$           .....................           $\frac{1}{50^2}<\frac{1}{49\cdot50}$$\Rightarrow\frac{1}{2^2}+...+\frac{1}{49^2}+\frac{1}{50^2}<\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}$Ta có: $\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}<1$$\Rightarrow\frac{1}{2^2}+...+\frac{1}{49^2}+\frac{1}{50^2}<1$$\Rightarrow A=\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{49^2}+\frac{1}{50^2}<2$

Nguyên · Answer

$A=\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{49^2}+\frac{1}{50^2}$$A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}+\frac{1}{50.51}$$A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}+\frac{1}{50}-\frac{1}{51}$$A=1-\frac{1}{51}$$A=\frac{50}{51}<1<2$ (đpcm)Câu trả lời: $A=\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{49^2}+\frac{1}{50^2}$$A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}+\frac{1}{50.51}$$A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}+\frac{1}{50}-\frac{1}{51}$$A=1-\frac{1}{51}$$A=\frac{50}{51}<1<2$ (đpcm)

Uchiha Sasuke · Answer

Ta có:$A=\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{49^2}+\frac{1}{50^2}<1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}$$A<1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}$$A<1+1-\frac{1}{50}$$A<1+\frac{49}{50}$$A<\frac{99}{50}<2$$A<2$=> điều cần chứng minhCâu trả lời: Ta có:$A=\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{49^2}+\frac{1}{50^2}<1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}$$A<1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}$$A<1+1-\frac{1}{50}$$A<1+\frac{49}{50}$$A<\frac{99}{50}<2$$A<2$=> điều cần chứng minh

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