Question

\(\frac{5}{1.3}+\frac{5}{3.5}+....+\frac{5}{99.101}\)

Answer 1

\(=\frac{2}{1.3}.\frac{5}{2}+\frac{2}{3.5}.\frac{5}{2}+...+\frac{2}{99.101}.\frac{5}{2}\)
\(=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}\)
\(=\frac{250}{101}\)

Answer 2

5/1.3 + 5/3.5 + ... + 5/99.101

= 5/2.(1 - 1/3 + 1/3 - 1/5 + ... + 1/99 - 1/101)

= 5/2.(1 - 1/101)

=5/2.100/101

= 250/101

Answer 3

\(\frac{5}{1.3}+\frac{5}{3.5}+..+\frac{5}{99.101}\)

=\(\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)

=\(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

= \(\frac{5}{2}\left(1-\frac{1}{101}\right)=\frac{5}{2}.\frac{100}{101}=\frac{500}{202}=\frac{250}{101}\)

Answer 4

\(\frac{5}{1.3}+\frac{5}{3.5}+..+\frac{5}{99.101}\)

\(=\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)

\(=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=\frac{5}{2}\left(1-\frac{1}{100}\right)\)

\(=\frac{5}{2}.\frac{99}{100}\)

\(=\frac{99}{40}\)