Câu hỏi của Nguyễn Dương

Question

Chứng tỏ rằng: $\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{3}{4}$

Trắng_CV · Accepted Answer

Làm theo cách của Trắng nha , $\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}$$\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{1}{2^2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}$$\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{1}{4}+\frac{1}{2}-\frac{1}{2019}$$\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{3}{4}-\frac{1}{2019}< \frac{3}{4}\left(Đpcm\right)$Câu trả lời: Làm theo cách của Trắng nha , $\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}$$\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{1}{2^2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}$$\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{1}{4}+\frac{1}{2}-\frac{1}{2019}$$\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{3}{4}-\frac{1}{2019}< \frac{3}{4}\left(Đpcm\right)$

kudo shinichi · Answer

Ta có:  $\frac{1}{2^2}=\frac{1}{2^2}$            $\frac{1}{3^2}< \frac{1}{2.3}$             ...................             $\frac{1}{2019^2}< \frac{1}{2018.2019}$$\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2019^2}< \frac{1}{2^2}+\frac{1}{2.3}+...+\frac{1}{2018.2019}$$=\frac{1}{2^2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2018}-\frac{1}{2019}$$=\frac{1}{4}+\frac{1}{2}-\frac{1}{2019}$$=\frac{1}{4}+\frac{2}{4}-\frac{1}{2019}$$=\frac{3}{4}-\frac{1}{2019}$$< \frac{3}{4}$$\Rightarrow$$\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2019^2}< \frac{3}{4}$                                              Điều phải chứng minhCâu trả lời: Ta có:  $\frac{1}{2^2}=\frac{1}{2^2}$            $\frac{1}{3^2}< \frac{1}{2.3}$             ...................             $\frac{1}{2019^2}< \frac{1}{2018.2019}$$\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2019^2}< \frac{1}{2^2}+\frac{1}{2.3}+...+\frac{1}{2018.2019}$$=\frac{1}{2^2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2018}-\frac{1}{2019}$$=\frac{1}{4}+\frac{1}{2}-\frac{1}{2019}$$=\frac{1}{4}+\frac{2}{4}-\frac{1}{2019}$$=\frac{3}{4}-\frac{1}{2019}$$< \frac{3}{4}$$\Rightarrow$$\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2019^2}< \frac{3}{4}$                                              Điều phải chứng minh

💛Linh_Ducle💛 · Answer

Đặt $A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}$Ta có:$\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2}$$\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3}$$\frac{1}{4^2}=\frac{1}{4.4}< \frac{1}{3.4}$....$\frac{1}{2019^2}=\frac{1}{2019.2019}< \frac{1}{2018.2019}$$\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}$$\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}$$\Rightarrow A< 1-\frac{1}{2019}$$\Rightarrow A< \frac{2018}{2019}$đến đây mới thấy mik sai ,xin lỗiCâu trả lời: Đặt $A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}$Ta có:$\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2}$$\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3}$$\frac{1}{4^2}=\frac{1}{4.4}< \frac{1}{3.4}$....$\frac{1}{2019^2}=\frac{1}{2019.2019}< \frac{1}{2018.2019}$$\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}$$\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}$$\Rightarrow A< 1-\frac{1}{2019}$$\Rightarrow A< \frac{2018}{2019}$đến đây mới thấy mik sai ,xin lỗi

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