\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=2.\frac{49}{100}\)
\(=\frac{49}{50}\)
= 2.(1/2.3 + 1/3.4 + ... + 1/99.100)
trong ngoac co cong thuc do, tim hieu di la lam dc
\(2\cdot A=2\cdot\left(\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{99\cdot100}\right)\)
\(2\cdot A=\frac{4}{2\cdot3}+\frac{4}{3\cdot4}+\frac{4}{4\cdot5}+...+\frac{4}{99\cdot100}\)
\(2\cdot A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(2\cdot A=1-\frac{1}{100}\)
\(2\cdot A=\frac{99}{100}\)
\(A=\frac{99}{100}:2=\frac{99}{200}\)
\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
Â
2/2.3+2/3.4+...+2/99.100
=2.98/2.100
=196/200
=49/50.
mình là đúng đó, cách mình tự tìm đó, rất hợp lí,nhanh,gọn.Bạn có thể áp dụng vào bài thi đó, k nhé ^_^
