Question

   a,    \(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+...........+\(\frac{1}{99.100}\)

   b,\(\frac{2}{3.5}\)+\(\frac{2}{5.7}\)+\(\frac{2}{7.9}\)+.........+\(\frac{2}{35.37}\)

   c,\(\frac{7}{7.9}\)+\(\frac{7}{9.11}\)+\(\frac{7}{11.13}\)+..........+\(\frac{7}{99.101}\)

Answer 1

a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}.\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)

b) \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{35.37}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{35}-\frac{1}{37}\)

\(=\frac{1}{3}-\frac{1}{37}=\frac{34}{111}\)

Answer 2

c) \(\frac{7}{7.9}+\frac{7}{9.11}+\frac{7}{11.13}+...+\frac{7}{99.101}\)

\(=\frac{7}{2}.\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{7}{2}.\left(\frac{1}{7}-\frac{1}{101}\right)=\frac{7}{2}\cdot\frac{94}{707}=\frac{47}{101}\)

Answer 3

a) 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/99.100

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/99 - 1/100

= 1 - 1/100

= 99/100

b) 2/3.5 + 2/5.7 + 2/7.9 + ... + 2/35.37

= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/35 - 1/37

= 1 - 1/37

= 36/37

c) 7/7.9 + 7/9.11 + 7/11.13 + ... + 7/99.101

= 7/2.(1/7 - 1/9 + 1/9 - 1/11 + 1/11 - 1/13 + ... + 1/99 - 1/101)

= 7/2.(1/7 - 1/101)

= 7/2.94/707

= 47/101