Câu hỏi của hyduyGF

Question

Bài 6:(0,5 điểm) Cho Chứng minh A < 2.

Dũng Nguyễn Đình · Accepted Answer

Ta có :$\frac{1}{1^2}< \frac{1}{1.2};\frac{1}{2^2}< \frac{1}{2.3};...;\frac{1}{50^2}< \frac{1}{49.50}$$\Leftrightarrow\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}=1-\frac{1}{50}< 1< 2$Vậy A < 2Câu trả lời: Ta có :$\frac{1}{1^2}< \frac{1}{1.2};\frac{1}{2^2}< \frac{1}{2.3};...;\frac{1}{50^2}< \frac{1}{49.50}$$\Leftrightarrow\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}{49.50}=1-\frac{1}{50}< 1< 2$Vậy A < 2

Lovers · Answer

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

hyduyGF · Answer

minh ko hieu\ Câu trả lời: minh ko hieu\

Khoá học trên OLM (olm.vn)

Khoá học trên OLM (olm.vn)