Câu hỏi của ĐINH THỊ DIỆU LINH

Question

A= $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}$chứng minh A <2giúp mình với phải gọi là trên khẩn cấp luôn gấp lắm

Phùng Minh Quân · 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}{50^2}< \frac{1}{49.50}$$\Rightarrow$$A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< \frac{1}{1^2}+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}$$\Rightarrow$$A< 1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}$$\Rightarrow$$A< 1+1-\frac{1}{50}< 1+1$$\Rightarrow$$A< 2$Vậy $A< 2$Chúc bạn học tốt ~ 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}{50^2}< \frac{1}{49.50}$$\Rightarrow$$A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< \frac{1}{1^2}+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}$$\Rightarrow$$A< 1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}$$\Rightarrow$$A< 1+1-\frac{1}{50}< 1+1$$\Rightarrow$$A< 2$Vậy $A< 2$Chúc bạn học tốt ~

Minh Nguyễn Cao · Answer

CM A < 2=> CM $\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1$Ta thấy: $\frac{1}{2^2}=\frac{1}{4}< \frac{1}{2}=\frac{1}{1.2}$ $\frac{1}{3^2}=\frac{1}{9}< \frac{1}{6}=\frac{1}{2.3}$Và cứ thế,.... $\frac{1}{50^2}=\frac{1}{2500}< \frac{1}{2450}=\frac{1}{49.50}$=> $\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}$ $=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1$=> $\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1$=> $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1+\frac{1}{1^2}=1+1=2$=>ĐPCMCâu trả lời: CM A < 2=> CM $\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1$Ta thấy: $\frac{1}{2^2}=\frac{1}{4}< \frac{1}{2}=\frac{1}{1.2}$ $\frac{1}{3^2}=\frac{1}{9}< \frac{1}{6}=\frac{1}{2.3}$Và cứ thế,.... $\frac{1}{50^2}=\frac{1}{2500}< \frac{1}{2450}=\frac{1}{49.50}$=> $\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}$ $=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1$=> $\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1$=> $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1+\frac{1}{1^2}=1+1=2$=>ĐPCM

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