Câu hỏi của Huỳnh Mun

Question

Cho A= 1/1^2+1/2^2+1/3^2+...+1/50^2. Chứng minh A<2

Đào Trọng Luân · Accepted Answer

$Cm:\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}$< 2Ta có: $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}$$\Rightarrow\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{49}{50}< 1< 2$=> A < 2tk nha mnCâu trả lời: $Cm:\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}$< 2Ta có: $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}$$\Rightarrow\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{49}{50}< 1< 2$=> A < 2tk nha mn

Lãng Tử Hào Hoa · Answer

Ta có: $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}$ $=1+\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)$ $=1+\left(\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{50.50}\right)< 1+\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{50.51}\right)$$\Rightarrow A< 1+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)$$\Rightarrow A< 1+\left(\frac{1}{2}-\frac{1}{51}\right)=1+\frac{49}{102}< 1+1=2$ (Đpcm)Câu trả lời: Ta có: $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}$ $=1+\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)$ $=1+\left(\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{50.50}\right)< 1+\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{50.51}\right)$$\Rightarrow A< 1+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)$$\Rightarrow A< 1+\left(\frac{1}{2}-\frac{1}{51}\right)=1+\frac{49}{102}< 1+1=2$ (Đpcm)

Dương Minh Anh · Answer

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

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