Câu hỏi của Kerry Meir

Question

Tính : ( 1/ 1 x 2 + 1/ 3 x 4 + ... + 1/ 99 x 100 ) - ( 1/51 + 1/52 + 1/100 )

Kiên-Messi-8A-Boy2k6 · Accepted Answer

Đặt $A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{99.100}$$\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}$$\Rightarrow A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)$$\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)$$\Rightarrow A=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)$$\Rightarrow A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}$$\Rightarrow A-\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)=0$$\Rightarrow\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)-\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)=0$Câu trả lời: Đặt $A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{99.100}$$\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}$$\Rightarrow A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)$$\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)$$\Rightarrow A=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)$$\Rightarrow A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}$$\Rightarrow A-\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)=0$$\Rightarrow\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)-\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)=0$

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