Question

a) 1/1.3+1/3.5+1/5.7+.....+1/99.101

b) 7/1.3+7/3.5+7/5.7+.....+7/99.101

 

 

Answer 1

a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}-...-\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}\)

\(=\frac{100}{101}\)

b) \(\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+...+\frac{7}{99.101}\)

\(=7.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{7}{5.7}+...+\frac{7}{99.101}\right)\)

\(=7.\frac{1}{7}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{7}{7}\left(1-\frac{1}{101}\right)\)

\(=\frac{100}{101}\)

Answer 2

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