a) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{99.100}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{99}-\dfrac{1}{100}\)
=\(\dfrac{1}{1}+0+0+...+0-\dfrac{1}{100}\)
=\(1-\dfrac{1}{100}\)
= \(\dfrac{99}{100}\)
a) 11.2+12.3+13.4+....+199.10011.2+12.3+13.4+....+199.100
= 11−12+12−13+13−14+....+199−110011−12+12−13+13−14+....+199−1100
=11+0+0+...+0−110011+0+0+...+0−1100
=1−11001−1100
= 99100