\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)
\(=2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
\(=2.\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=1-\frac{1}{100}\Rightarrowđpcm\)
Ta có :
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}< 1\)\(\left(đpcm\right)\)
=1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101=1-1/101=100/101<1
2 câu cuối pn sửa lại là 101 nha , mình sr nhá
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)\(=\frac{1}{1}-\frac{1}{101}\)\(=\frac{100}{101}\)\(Mà\frac{100}{101}< 1\)\(Nên\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}< 1\)