\(a,=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
\(b,=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)
a,\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}=\frac{2}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)=\frac{2}{2}.\left(\frac{1}{1}-\frac{1}{100}\right)=1.\frac{99}{100}=\frac{99}{100}\)
a, 2/1.3 + 2/3.5 +2/5.7 +...+2/99.101
= 1.(1-1/3 +1/3-1/5+1/5-1/7+....+1/99-1/101)
= 1.(1-1/101)
= 1.100/101
= 100/101
b, 5/1.3 + 5/3.5 + 5/5.7 +...+ 5/99.101
= 5/2 .( 1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101)
= 5/2. (1-1/101)
= 5/2 . 100/101
= 250/101
a) \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
= \(2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
= \(2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
= \(2.\left(\frac{1}{1}-\frac{1}{101}\right)\)
= \(2.\left(\frac{101}{101}-\frac{1}{101}\right)\)
= \(2.\frac{100}{101}\)
= \(\frac{200}{101}\)
b) \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
= \(5.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
= \(5.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99.101}\right)\)
= \(5.\left(\frac{1}{1}-\frac{1}{101}\right)\)
= \(5.\left(\frac{101}{101}-\frac{1}{101}\right)\)
= \(5.\frac{100}{101}\)
= \(\frac{500}{101}\)